Physics (Volume 2) | Grade 12
Competency Focused Practice Questions | Physics | Grade 12
PREFACE
Assessments are an important tool that help gauge learning. They provide valuable feedback
about the effectiveness of instructional methods; about what students have actually understood
and also provide actionable insights. The National Education Policy, 2020 has outlined the
importance of competency‐based assessments in classrooms as a means to reform curriculum
and pedagogical methodologies. The policy emphasizes on the development of higher order skills
such as analysis, critical thinking and problem solving through classroom instructions and aligned
assessments.
Central Board of Secondary Education (CBSE) has been collaborating with Educational Initiatives
(Ei) in the area of assessment. Through resources like the Essential Concepts document and
A‐Question‐A‐Day (AQAD), high quality questions and concepts critical to learning have been
shared with schools and teachers.
Continuing with the vision to ensure that every student is learning with understanding, Question
Booklets have been created for subjects for Grade 10th and 12th. These booklets contain
competency‐based items, designed specifically to test conceptual understanding and application
of concepts.
Process of creating competency‐based items
All items in these booklets are aligned to the NCERT curriculum and have been created keeping
in mind the learning outcomes that are important for students to understand and master. Items
are a mix of Free Response Questions (FRQs) and Multiple‐Choice Questions (MCQs). In case of
MCQs, the options (correct answer and distractors) are specifically created to test for
understanding and capturing specific errors/misconceptions that students may harbour. Each
incorrect option can thereby inform teachers on specific gaps that may exist in student learning.
In case of subjective questions, each question also has a detailed scoring rubric to guide
evaluation of students’ responses.
Each item has been reviewed by experts, to check for appropriateness of the item, validity of the
item, conceptual correctness, language accuracy and other nuances.
How can these item booklets be used?
There are 179 questions in this booklet.
The purpose of these item booklets is to provide samples of high‐quality competency‐based items
to teachers. The items can be used to–
● get an understanding of what good competency‐based questions could look like
● give exposure to students to competency‐based items
● assist in classroom teaching and learning
● get inspiration to create more such competency‐based items
Students can also use this document to understand different kinds of questions and practice
specific concepts and competencies. There will be further additions in the future to provide
competency focused questions on all chapters.
Please write back to us to give your feedback.
Team CBSE
Competency Focused Practice Questions | Physics | Grade 12
CONTENTS
1.
ELECTROSTATICS........................................................................................................................... 1
Answer Key & Marking Scheme ................................................................................................. 12
2.
CURRENT ELECTRICITY ................................................................................................................ 26
Answer Key & Marking Scheme ................................................................................................. 33
3.
MAGNETIC EFFECTS OF CURRENT AND MAGNETISM ................................................................ 41
Answer Key & Answer Scheme................................................................................................... 57
4.
ELECTROMAGNETIC INDUCTION AND ALTERNATING CURRENT (AC) ........................................ 70
Answer Key & Marking Scheme ................................................................................................. 82
5.
ATOMS AND NUCLEI ................................................................................................................... 93
Answer Key & Marking Scheme ................................................................................................. 97
6.
DUAL NATURE OF RADIATION AND MATTER ............................................................................. 99
Answer Key & Marking Scheme ............................................................................................... 102
7.
ELECTROMAGNETIC WAVES ..................................................................................................... 104
Answer Key & Marking Scheme ............................................................................................... 107
8.
OPTICS....................................................................................................................................... 110
Answer Key & Marking Scheme ............................................................................................... 121
Competency Focused Practice Questions | Physics | Grade 12
1. ELECTROSTATICS
Q. No
Question
Marks
Multiple Choice Questions
Q.1
The capacitance of a capacitor is Co. It is connected to a battery of voltage V
which charges the capacitor. With the capacitor still connected to the battery,
a slab of dielectric material is introduced between the plates of the capacitor.
1
Which of the following explains the effect of the dielectric slab in the above
situation?
Q.2
A.
The electric field between the plates of the capacitor rises.
B.
The potential difference between the plates falls.
C.
The total charge on the capacitor increases.
D.
The ability of the capacitor to store charge decreases.
The image below shows two examples of electric field lines.
1
Which of the following statements is true?
A.
The electric fields in both I and II arise due to a single positive point charge
located somewhere on the left.
B.
The electric fields in both I and II can be created by negative charges
located somewhere on the left and positive charges somewhere on the
right.
C.
The electric field in I is the same everywhere but the electric field in II
becomes stronger as we move from left to right.
D.
As you move from left to right, the electric fields in both I and II become
stronger.
Competency Focused Practice Questions | Physics | Grade 12
1
Q.3
In a given region, electric potential varies with position as V(x) = 3 + 2x2.
1
Identify which of the following statements is correct.
Q.4
A.
Potential difference between the two points x = 2 and x = ‐2 is 2 V.
B.
A charge of 1 C placed at x = 2 experiences a force of 6 N.
C.
The force experienced by the above charge is along +x ‐ axis.
D.
The electric field in the given region is non‐uniform along x ‐ axis.
A parallel plate capacitor is charged to a potential difference V. Two protons P
and Q are placed at the two locations inside the capacitor as shown.
1
Which one of the statements is correct?
A. The forces on the two protons are identical.
B. The force on proton P near the positive plate is more than the force on
proton Q.
C. The force on proton Q near the negative plate is more than the force on
proton P.
D. The forces on both the protons are zero.
Q.5
In a given region of an electric field, there is no charge present. A closed container
is placed in this region of the electric field.
1
What is the requirement for the total flux through the closed container to be
zero?
A.
The field must be uniform.
B.
The container must be symmetric.
C.
The container must be oriented in a particular direction.
D.
There is no such requirement. The total flux through the container is zero
no matter what.
Competency Focused Practice Questions | Physics | Grade 12
2
Q.6
A uniform electric field of 5000 V/m exists in a certain region. What volume of
this space will contain energy equal to 10‐6 J? (Take ∈o = 8.8 x10‐12 SI units)
A.
9 m3
B.
9 x 10‐3 m3
C.
9 x 10‐6 m3
D.
9 x 103 m3
1
Free Response Question/Subjective Question
Q.7
Given are three different square arrangements of charges.
4
(a) Draw relevant vector diagrams to represent the resultant electric field ER at
the center of the square in each case.
(b) Identify the arrangement in which the resultant electric field is the smallest
at the center of the square.
Q.8
The figure below shows an arrangement of four charges along with some electric
field lines drawn between the charges.
4
(a) Identify three things that are incorrect in this figure.
(b) Draw a correct diagram representing the electric field lines for this system of
charges.
Competency Focused Practice Questions | Physics | Grade 12
3
Q.9
In the case of a spherical charged conductor of radius R, hollow or solid, the
electric potential is constant and maximum inside and on the surface.
3
On the outside, the electric potential varies as,
where r is the distance from the center of the sphere.
Q.10
(a)
Represent the variation of electric potential due to a charged sphere with
distance graphically.
(b)
How does the electric field intensity due to a charged sphere vary with
distance r from the center in the above case?
(c)
Represent the variation of electric field intensity due to a charged sphere
with distance graphically.
For a given charge distribution, an equipotential surface is the locus of all points
having the same potential.
3
Draw two equipotential surfaces for each of the following:
Q.11
(a)
a uniform electric field
(b)
a point charge
(c)
an infinite straight line of charge
Study the graph between electric field intensity E versus the distance r.
(a)
Describe the nature of variation of electric field intensity between r = 0 and
r ≤ a.
(b)
Describe the nature of variation of electric field intensity for r > a.
(c)
Give an example of the body of charge distribution that can exhibit the
above studied electric field distribution.
Competency Focused Practice Questions | Physics | Grade 12
3
4
Q.12
Given is a pair of parallel charged metal plates in the arrangement as shown.
3
(a) Sketch the electric field lines between the plates in I and II.
(b) Mention whether net electric field intensity is zero or non‐zero in each
case.
Q.13
The dielectric strength of a medium is the minimum electric field required to
cause ionization of the medium. For air, this value is taken as 3 million V/m.
2
With this information, find out if a metal sphere of radius 1 cm, surrounded by
air, can hold a charge of 1 coulomb.
Q.14
(a) If electric field strength at a point is zero at a given point, then what can you
say about the electric potential at that point? Explain.
2
(b) In the two instances below, state whether electric field intensity and electric
potential are zero or non‐zero at the mid‐point joining the two‐point charges.
Competency Focused Practice Questions | Physics | Grade 12
5
Q.15
A metal sphere A of radius a is charged to a potential V. It is enclosed inside a
hollow metal shell B of radius b.
2
What will be the potential of A if it is connected to shell B by a conducting wire?
Show the working.
Q.16
Two hollow spherical conductors of different sizes are charged positively. The
smaller one is at 50 V and the larger one is at 100 V.
3
Suggest a method in which the two conductors can be arranged so that the
charge flows from the smaller to the bigger conductor when connected by a
wire.
Give a mathematical working to justify the arrangement.
Q.17
The figure below shows lines of constant potentials in a region in which an
electric field is present.
2
Using the relation between potential difference and electric field intensity, find
which of the three points, A, B or C will have the greatest electric field intensity.
Competency Focused Practice Questions | Physics | Grade 12
6
Q.18
The capacitance of an infinite parallel capacitor without any dielectric between
the plates is C. It is then half‐filled with a dielectric medium of K = 4 as in Fig (a)
and then as in Fig (b).
3
Show that capacitance in Fig (a) becomes 5C and in Fig (b) the capacitance
becomes (4/5)C due to the introduction of the dielectric medium.
Q.19
Q.20
Two small balls, each with a charge Q, hang from the same point by insulating
strings of length L from a fixed support. Consider the setup in a region of zero
gravity and in equilibrium.
(a)
What will be the angle between the two strings?
(b)
What will be the tension in each of the strings?
A conducting wire connects two charged conducting spheres such that they
attain equilibrium with respect to each other. The distance of separation
between the two spheres is very large as compared to either of their radii.
2
2
Find the ratio of the magnitudes of the electric fields at the surfaces of the two
spheres.
Competency Focused Practice Questions | Physics | Grade 12
7
Q.21
A parallel plate capacitor of capacitance C is charged to a potential V by a
battery. Q is the charge stored on the capacitor. Without disconnecting the
battery, the plates of the capacitor are pulled apart to a larger distance of
separation.
5
What changes will occur in each of the following quantities? Will they increase,
decrease or remain the same? Give an explanation in each case.
(a) Capacitance
(b) Charge
(c) Potential difference
(d) Electric field
(e) Energy stored in the capacitor
Q.22
A parallel plate capacitor C with a dielectric in between the plates is charged to a
potential V by connecting it to a battery. The capacitor is then isolated.
3
If the dielectric is withdrawn from the capacitor,
Q.23
(a)
Will the energy stored in the capacitor increase or decrease? What causes
the change in energy?
(b)
Will the potential difference across the capacitor plates increase or
decrease? Give an explanation.
A small ball of mass 2 x 10‐16 kg carrying a charge q = ‐ 2 μC is fired from the
positive plate of the capacitor towards the negative plate with a speed of 3 x
106 m/s.
3
Will the ball strike the negative plate? Give mathematical working for the answer.
Competency Focused Practice Questions | Physics | Grade 12
8
Q.24
An isolated capacitor of unknown capacitance C is charged to a potential
difference V. It is then connected in parallel to an uncharged capacitor of
capacitance Co such that the potential difference across the combination
becomes V/3.
2
Determine the unknown capacitance C.
Q.25
P and Q are two concentric conducting hollow shells connected to each other
by a conducting wire.
2
If a total charge of 10 C is given to this system of two concentric shells, how much
charge will settle on each shell? Explain.
Q.26
A thin rod of length L of the uniform mass density of μ kg/m and linear charge
density λ C/m is initially at rest.
3
A uniform electric field E is applied in the direction perpendicular to the length
of the rod. The rod begins to move in the direction of the applied electric field.
Determine the speed of the rod after it travels through 1 m of distance. Ignore
all external forces except the electrostatic forces.
Competency Focused Practice Questions | Physics | Grade 12
9
Q.27
A thundercloud carries a charge of +50 C at a height of 4000 m and a charge of
‐50 C at a height of 2000 m from the ground. An airplane crosses through the
charged thundercloud at a height of 3000 m from the ground.
2
Find the magnitude and the direction of the electric field acting on the airplane
as its crosses through the charged‐up thundercloud.
Q.28
Given below is the representation of equipotential lines in a given electric field
region. Determine the strength and direction of electric field vectors at points 1
and 2.
3
Q.29
Two identical small metal blocks each of mass m are held to each other by a
metallic spring of force constant k. The block‐spring system is placed on a
frictionless surface and the spring has an unstretched length L1 as in Fig (i).
2
A total charge Q is slowly given to the block‐spring system resulting in the spring
getting stretched to an equilibrium length L2 as in Fig (ii).
Assume that the entire charge gets distributed only on the two blocks and
consider the two blocks as point charges. Determine the charge Q.
Competency Focused Practice Questions | Physics | Grade 12
10
Q.30
Q.31
The diagram below shows an arrangement of a set of equipotential lines in a
given region. Each line is marked with the potential it represents. The
background gridlines are squares of each side equal to 1 cm.
(a)
Determine electric field at point P.
(b)
Draw 6 field lines to represent the electric field in this region.
The area of each of the plates of a parallel plate air capacitor is 7 cm2.
2
3
(a) Determine the maximum charge this capacitor can store without breakdown.
(b) A material of dielectric constant 2 and dielectric strength 15 x 106 V/m is
inserted into the capacitor. Find the percentage change in the maximum charge
that can be stored in the capacitor with the dielectric material.
(Take Dielectric strength of air = 3 x 106 V/m, ∈o = 8.8 x10‐12 C2/Nm2)
Competency Focused Practice Questions | Physics | Grade 12
11
Answer Key & Marking Scheme
Q. No
Answers
Marks
Q.1
C. The total charge on the capacitor increases.
1
Q.2
C. The electric field in I is the same everywhere but the electric field in II
becomes stronger as we move from left to right.
1
Q.3
D. The electric field in the given region is non‐uniform along x ‐ axis.
1
Q.4
A. The forces on the two protons are identical.
1
Q.5
D. There is no such requirement. The total flux through the container is zero no
matter what.
1
Q.6
B. 9 x 10‐3 m3
1
Q.7
(a) 1 mark each for the correct vector representation of the electric field in the
three cases:
4
Competency Focused Practice Questions | Physics | Grade 12
12
(b) ER vector in X is the smallest in length.
Q.8
(a) 1 mark for each correct point:
4
‐ Electric field lines cross each other as shown at point P.
‐ Number of field lines that end on the negative charges is not proportional to
their charges.
‐ The field lines drawn between +4q and –q are shown as parallel and
equidistant.
(b) The correct representation:
[1 mark for the correct representation of the electric field lines]
Competency Focused Practice Questions | Physics | Grade 12
13
Q.9
(a)
3
[1 mark for the correct graph]
(b) Electric field intensity is zero inside the charged sphere. Maximum at the
surface. On the outside of the sphere, it falls as E proportional to 1/r2
[1 mark for the correct statement of variation of E with r]
(c)
[1 mark for the correct graph]
Q.10
(a) A uniform electric field
Competency Focused Practice Questions | Physics | Grade 12
3
14
[1 mark for the correct drawing of two equipotential surfaces]
(b) A point charge
[1 mark for the correct drawing of two equipotential surfaces]
(c) An infinite straight line of charge
[1 mark for the correct drawing of two equipotential surfaces]
Q.11
E = 0 at r = 0.
3
Between r = 0 and r < a, the electric field intensity increases
linearly.E is maximum at r = a.
[0.5 mark for each point]
(b)For r > a, the electric field intensity decreases as 1/r2.
[0.5 mark for this point]
(c) A uniformly charged insulating solid sphere.
[1 mark for the example]
Competency Focused Practice Questions | Physics | Grade 12
15
Q.12
(a)
3
[1 mark for correct representation of electric field inside metal]
[1 mark for correct representation of electric field inside dielectric]
(b) The net electric field intensity inside a metal is zero.
The net electric field intensity inside a dielectric is non‐zero.
[0.5 mark for each point]
Q.13 Electric field on the surface of a metal sphere:
2
[1 mark for the correct formula and calculation]
As this field is much higher than the dielectric strength of the surrounding air
(=3 x 106 V/m), a sphere of radius 1 cm CANNOT hold 1 C of charge in air.
[1 mark for the correct interpretation and conclusion]
Q.14
(a) As
2
If E =0, at a given point, then
i.e., V = 0 or constant at that point.
[1 mark for correct explanation]
(b) At mid‐point P in Fig I, E is zero, but V is non‐zero.
At mid‐point P in Fig II, E is non‐zero, but V adds up to zero.
[0.5 mark for each point]
Competency Focused Practice Questions | Physics | Grade 12
16
Q.15
Potential on the surface of sphere A:
2
On connecting sphere, A with shell B, the charge will now reside on the surface
of the shell.
Potential on the surface of shell B:
[0.5 mark for each correct formula]
Comparing the two equations:
Since sphere A is inside the shell B, so its potential now becomes the same as
that of B, that is,
[1 mark for the correct explanation]
Q.16 The smaller conductor can be enclosed inside the larger conductor and the two
are connected by a conducting wire.
3
[1 mark for the correct arrangement of the two conductors]
Potential on the surface of the smaller conductor placed inside the bigger
conductor will be
Vi = 50 V + 100 V = 150 V
[0.5 mark for the correct formula]
Potential on the surface of the outer conductor will be:
[1 mark for the correct formula and calculation]
Hence when the inner conductor is connected by a conducting wire to the outer
conductor, the charge flows from the higher potential surface of the inner
conductor to the lower potential surface of the outer conductor.
[0.5 mark for the correct conclusion]
Competency Focused Practice Questions | Physics | Grade 12
17
Q.17
Relation between E and V is
2
Here ΔV between successive lines is constant, that is 10V.
[1 mark for correct formula]
Therefore,
The smaller the distance of separation between two successive lines, the more
is the electric field.
So point B has the greatest electric field intensity.
[1 mark for correct explanation]
Q.18
In Fig (a), the two capacitors are in parallel combination.
3
So the equivalent capacitance = C + KC = C [1+K] = 5C
[0.5 mark for correct identification of capacitor combination]
[1 mark for correct calculation of equivalent capacitance]
In Fig (b), the two capacitors are in a series combination.
Equivalent capacitance =
[0.5 mark for correct identification of capacitor combination]
[1 mark for correct calculation of equivalent capacitance]
Q.19
(a) The angle between the two strings will be 180o.
2
[1 mark for the correct angle]
(b) Tension in each string will be equal to the electrostatic force of repulsion
between the two charged balls.
[1 mark for the correct formula and substitution]
Competency Focused Practice Questions | Physics | Grade 12
18
Q.20
At equilibrium,
2
the potential on the surface of a larger sphere = potential on the surface of a
smaller sphere.
So,
[1 mark for the correct ratio of q1 to q2]
Since the two charges are very far from each other, the electric fields on the
surfaces of the two spheres will be:
The ratio of the electric fields is,
[1 mark for the correct ratio of E1 to E2]
Q.21
(a) Capacitance decreases.
5
Capacitance is inversely proportional to the distance of separation.
[0.5 mark for correct change]
[0.5 mark for correct explanation]
(b) Charge decreases.
From Q=CV, C decreases and V remains the same, so Q decreases.
[0.5 mark for correct change]
[0.5 mark for correct explanation]
(c) Potential difference remains the same
As the capacitor is connected to the battery, the potential V of the capacitor will
remain the same as that of the battery.
[0.5 mark for correct change]
[0.5 mark for correct explanation]
(d) Electric field decreases.
Competency Focused Practice Questions | Physics | Grade 12
19
E due to a plane sheet of charge = σ/∈o is independent of the distance from the
sheet. But charge density σ on the plate decreases, so E decreases.
OR:
Alternatively,
As E = V/d = Q/Cd = Q/∈oA
Since Q decreases, E also decreases.
[0.5 mark for correct change]
[0.5 mark for correct explanation]
(e) Energy stored in the capacitor decreases.
Energy stored is proportional to both Q and V. Charge Q decreases but potential
V is constant.
[0.5 mark for correct change]
[0.5 mark for correct explanation]
Q.22
(a) The energy of the capacitor will increase.
3
The work done on the capacitor while removing the dielectric results in an
increase in energy stored in the capacitor.
(or)
Energy of the capacitor after the dielectric is removed: E' = Q2/2C'
The charge on the plates remains the same as the capacitor is isolated. But since
the capacitance decreases when the dielectric is removed, the energy of the
capacitor will increase.
[0.5 marks for the correct change]
[1 mark for correct explanation]
(b) Potential difference will increase.
Potential difference after the dielectric is removed is V’ = Q/C’
where C’ is capacitance without the dielectric.
Since C’ < C and the charge Q remains the same in an isolated capacitor, the
potential difference V’ > V.
[0.5 marks for the correct change]
[1 mark for correct explanation]
Competency Focused Practice Questions | Physics | Grade 12
20
Q.23
Kinetic energy of the ball
3
KE = 0.5 x 2x10‐16 x (3x106)2 = 9 x 10‐4 J
[1 mark for the correct calculation of KE value]
For the ball to move through the electric field, work to be done against the
electric field is
W = U = qV = 2 x 10‐6 x 50 = 10‐4 J
[1 mark for the correct calculation of electrostatic energy required]
Since the KE of the ball > Energy required to move through the field between
the plates of the capacitor, the ball will strike the negative plate.
[1 mark for the correct conclusion]
Q.24
Charge on unknown capacitor = Charge on the combination
2
[1 mark for a correct statement of equality of charge on the two capacitors]
CV = (C parallel Co)eqv .V/3
CV= (C + Co).V/3
C = Co/2
[1 mark for correct calculation]
Q.25
There will be no charge on the inner sphere P. The entire charge of 10 C will be
on the outer shell Q.
2
[1 mark for the correct charge distribution]
In case some charge does appear on the inner sphere P, it creates an electric
field between the inner and the outer shell causing the charge to move to the
outer shell. So only the outer sphere gathers all the charge.
OR
In case some charge does appear on the inner sphere P, the inner shell will be
at a higher potential than the outer shell Q. This will cause the entire charge to
move from P to Q, so that potentials on both the shells becomes the same.
[1 mark for the correct charge explanation]
Competency Focused Practice Questions | Physics | Grade 12
21
Q.26
Take potential V = 0 at initial point.
3
After distance d,
Potential V = ‐ Ed
[1 mark for correct potential difference]
The electrostatic potential energy of the rod after traveling through a distance
d,
Uf = charge x potential difference = ‐ λL.Ed
[1 mark for correct final energy]
Comparing initial and final energies of the rod,
OR
Alternatively,
F = qE = ma
where q = λL and m = μL
substituting, for q and m,
acceleration a = Eλ/μ.
Using the equation of motion,
[1 mark for correct final result]
Competency Focused Practice Questions | Physics | Grade 12
22
Q.27
Electric field due to +50 C above the airplane:
2
= 4.5 x 105 N/C, acting downwards.
[0.5 mark for the correct value of electric field]
Electric field due to ‐50 C above the airplane:
= 4.5 x 105 N/C, acting downwards.
[0.5 mark for the correct value of electric field]
So, the total electric field acting on the airplane = 4.5 x 105 N/C + 4.5 x 105 N/C
= 9 x 105 N/C, acting downwards.
[1 mark for the correct value of the final electric field]
Q.28
At point 1:
3
40 V change over a distance 2 cm.
So electric field at point 1:
E1 = 40/0.02 = 2000 V/m
The direction of E1 will be along +x axis.
[1 mark for correct calculation of the E1 value]
[0.5 mark for a correct direction of E1]
At point 2:
40 V change over a distance of 4 cm.
So electric field at point 2:
E2 = 40/0.04 = 1000 V/m
The direction of E2 will be along +y axis.
[1 mark for correct calculation of the E2 value]
[0.5 mark for a correct direction of E2]
Competency Focused Practice Questions | Physics | Grade 12
23
Q.29
Force of electrostatic repulsion Fe between the two blocks is balanced by
restoring force Fr due to spring,
2
[0.5 mark each for the correct formula of the forces]
At equilibrium,
Fe = F r
Solving for Q,
[1 mark for correct final result]
Q.30
(a) Electric field at P =
2
= ‐ 100 N/C , directed downwards.
[0.5 mark for the correct formula]
[0.5 mark for the correct result]
(b)
[1 mark for the correct drawing of electric fields lines]
Competency Focused Practice Questions | Physics | Grade 12
24
Q.31
(a) Qmax = C Vmax
3
Qmax = 1 x 8.8 x10‐12 x 7 x 10‐4 x 3 x 106 = 184.8 x 10‐10 = 18.48 x 10‐9 = 18.48 nC
[0.5 mark for the correct formula]
[1 mark for the correct result]
(b) Qmax = 2 x 8.8 x10‐12 x 7 x 10‐4 x 15 x 106 = 1848 x 10‐10 = 184.8 x 10‐9 =
184.8 nC
The change in maximum charge that can be stored = (change in charge /
original charge) % = 900%
[1 mark for the correct calculation and result]
[0.5 mark for the correct calculation of % change]
Competency Focused Practice Questions | Physics | Grade 12
25
2. CURRENT ELECTRICITY
Q. No
Question
Marks
Multiple Choice Question
Q.32
If the voltmeter in the given circuit reads 8 V, what is the resistance of the
voltmeter?
1
A. 70/4119 Ω
B. 70 Ω
C. 420 Ω
D. 4200 Ω
Competency Focused Practice Questions | Physics | Grade 12
26
Q.33
Which combination of the three resistors will dissipate maximum power from a
10 V battery?
1
I
Q.34
A.
II
B.
III
C.
IV
If the ammeter reading in the given circuit is zero, find the value of the
resistance R.
A.
50 ohm
B.
100 ohm
C.
150 ohm
Competency Focused Practice Questions | Physics | Grade 12
1
27
D.
Q.35
200 ohm
Resistances 10 Ω and R Ω are connected in the two gaps of a metre bridge. The
null point is obtained at 30 cm on the metre scale from the zero end.
1
What resistance must be connected in parallel to R so that the null point is at
the mid‐point of the metre wire?
A.
10 Ω
B.
70/3 Ω
C.
70/4 Ω
D.
3/70 Ω
Free Response Question/ Subjective Question
Q.36
Accelerated alpha particles, constituting a beam current of 0.64 mA, are
released by a cyclotron onto a target. Find the number of alpha particles that
strike the target in one second.
2
Q.37
Alloys like Nichrome, whose temperature coefficient of resistance is
~ 0.4 x 10‐3 /oC at 20 oC, are used widely in the making of wire‐bound standard
resistors as their resistance changes very little with changes in temperature.
2
(i) Find the temperature at which the resistance of a given Nichrome wire would
be reduced to half its value at 20 oC.
(ii) State an anomaly regarding the temperature value obtained in (i).
Q.38
A battery of 6 V drives a current of 60 mA through an electric lamp. Another
battery of 10 V drives a current of 70 mA through the same lamp. Is the lamp an
ohmic device? Explain.
2
Q.39
A dimmer is a device that is connected to a light bulb and is used to lower the
brightness of the light from the bulb. It introduces a resistance in series with the
bulb for this purpose. A 44 W bulb in a household circuit is connected across a
220 V power supply.
3
(a) How much current does the bulb draw initially without the dimmer in action?
(b) What resistance in series would the dimmer introduce in order to drive a
current of 0.1 A through this bulb?
Competency Focused Practice Questions | Physics | Grade 12
28
Q.40
Eva needs a 3 V power supply for her electronic circuit experiment. There is a
10 V battery in the lab that she can use. Eva decides to make a voltage divider
using this battery and two different resistors, the larger value resistor being 300
ohm.
2
What value of the smaller resistor should Eva use to make the desired voltage
divider?
Q.41
A flashlight uses two batteries, each of emf 2 V and internal resistance 0.1 ohm,
in series. The flashlight bulb has a resistance of 10 ohm.
4
(a) What is the current drawn by the flashlight bulb?
(b) How much power is dissipated through the flashlight bulb?
(c) If the two batteries have zero internal resistances, will the power dissipated
through the flashlight bulb be more or less? Calculate the difference.
Q.42
In order to discourage the use of personal electrical appliances by the guests in
the rooms of a motel in a small town, each 220 V socket in the room is wired
with a circuit breaker on a 5 A line.
2
Sam wishes to iron his clothes using his 1200 W, 220 V portable steam iron box
in his motel room.
(a) Will Sam be able to use his iron box without tripping the circuit breaker?
Explain.
(b) Find the resistance of his iron box.
Competency Focused Practice Questions | Physics | Grade 12
29
Q.43
Tim is a music enthusiast who wishes to create good sound effects for his stereo
system using his two sets of speakers at home.
3
The first set consists of two speakers of resistance 10 Ω each.
The second set consists of two speakers with resistances 5 Ω and 10 Ω
respectively.
Initially, he connects the first set of two 10 Ω speakers in series to the 10 V
stereo output.
Later, he connects the second set of speakers such that each of them is parallel
to the first set of speakers.
(a) What is the current through each of the 10 Ω speakers BEFORE the second
set of speakers were connected?
(b) What is the new current through each of the 10 Ω speakers AFTER the
second set of speakers are connected?
(c) If the loudness of the music is directly proportional to the amount of power
used by the speaker, how has the loudness of the first set of speakers changed
due to the introduction of the second set of speakers?
(d) What is the impact of connecting the second set of speakers in the circuit
here?
Q.44
Rene uses a portable electric lamp to light up her keyboard when working on
her computer at night. The lamp consists of two small bulbs each of resistance
2 ohm connected in series and is powered by a 20 V battery.
2
(a) How much current is drawn by the portable lamp?
(b) Each of the two bulbs in the portable lamp is rated for continuous use of 10
hours only before they burn out. If the battery can supply total energy of 4000
kJ, which of the two occurs first: the lamp burns out or the battery drains out?
Show the working.
Q.45
In the given network of resistors, the 16‐ohm resistor burns out as soon as it
begins to dissipate 100 W of power.
Competency Focused Practice Questions | Physics | Grade 12
3
30
What MAXIMUM voltage can be applied across terminals A and B, in order to
avoid the burn out of the 16‐ohm resistor?
Q.46
Determine the equivalent resistance of the following combination of resistors.
Show the working (include diagrams if required).
2
.
Q.47
A Wheatstone’s bridge employs 5 resistors as shown. Find the value of n that
will ensure that the bridge is always in a balanced condition, irrespective of the
value of r.
Competency Focused Practice Questions | Physics | Grade 12
2
31
Q.48
Determine the value of R in the given network for each of the conditions
separately.
4
i. If VB – VD = 8 V
ii. If VA ‐ VC = 8 V
Competency Focused Practice Questions | Physics | Grade 12
32
Answer Key & Marking Scheme
Q.No
Answers
Marks
Q.32
C. 420 Ω
1
Q.33
A. I
1
Q.34
C. 150 ohm
1
Q.35
C. 70/4 Ω
1
Q.36
Current I = nQ/t = n x 2e/1
2
[1 mark for the correct formula]
[1 mark for the correct calculations]
Q.37
(i) Using
2
R = Ro (1 + α ΔT)
Ro /2 = Ro (1 + α ΔT)
ΔT = ‐ 1250 oC
T – To = ‐1250 oC
T = ‐ 1230 oC
[0.5 marks for the correct formula]
[0.5 marks for the correct calculations]
(ii) Anomaly: Since the T = ‐ 1230 oC, is below 0 K or ‐273 oC, it is physically
impossible to arrive at this temperature.
[1 mark for the correct point]
Competency Focused Practice Questions | Physics | Grade 12
33
Q.38
It’s a non‐Ohmic device.
2
[0.5 mark for correct conclusion]
Explanation:
Voltage change: v2/v1 = 10/6 = 0.6
Current change: I2/I1 = 70/60 = 1.16
When the voltage rises by a factor of 0.6, the corresponding current rises by the
factor, 1.16.
Since the proportion change in voltage and the current is not the same, the
electric lamp is a NON‐Ohmic device.
[1 mark for the correct calculations of voltage and current change]
[0.5 mark for the correct explanation of the conclusion]
OR
Resistance of the circuit in the first case= V/I = 6/60 x 10‐3 = 100 ohm
Resistance of the circuit in the second case= V/I = 10/70 x 10‐3 = 1000/7 = 142.8
ohm.
The resistance represents the slope of V‐I graph for the circuit and from the
above calculations, it doesn’t remain constant.
It implies that V‐I graph is not a straight line.
Hence the lamp used in the circuit is a non‐ohmic device.
[1 mark for the correct calculations of resistance values]
[0.5 mark for the correct explanation of the conclusion]
Q.39
(a) Initially, without any dimmer in action, current through the bulb:
3
P = VI
I = P/V = 44/220 = 1/5 = 0.2 A
[1 mark for the correct calculation of the current]
(b) Resistance of the bulb:
P= V2 /R1
R = 220 x 220 / 44 = 1100 ohm
With the introduction of the resistance in series by the dimmer,
V=IR
R = 220 /(0.1) = 2200 ohm
Competency Focused Practice Questions | Physics | Grade 12
34
R = R1 + R 2
where R2 is the resistance in series introduced by the dimmer
So R2 = 2200 – 1100 = 1100 ohm.
[1 mark for the correct value of resistance R1]
[1 mark for the correct value of resistance R2]
Q.40
Two resistors R1 (= 300 ohm) and R2 are connected in series for making the
voltage divider.
2
R1 = 300 ohm,
v1 = I R1
7 = 300.I ……(1)
[0.5 mark each for writing equation (1)]
Across the combination of two resistors,
10 = (R1 + R2) I …..(2)
Solving for smaller resistor R2
R2 = 900/7 ohm = 128.6 ohm
[0.5 mark each for writing equation (2)]
[1 mark for calculation and correct result]
Q.41
(a) Total voltage across the two batteries in series = 2 + 2 = 4 V
4
Total resistance in circuit = 0.1 + 0.1 + 10 = 10.2 ohm
Current through the flashlight bulb = I = V/R = 4/10.2 A
[1 mark for the correct value of the current]
(b) Power dissipated through the flashlight bulb = I2 R = (4/10.2)2 x 10 = 1.52 W
[1 mark for the correct value of the current]
(c) If the two batteries have zero internal resistances, the power dissipated
through the flashlight bulb will increase.
The current through the flashlight bulb, I = V/R = 4/10 = 0.4 A
Power dissipated through the flashlight bulb = I2 R = (0.4)2 x 10 = 1.6 W
The difference = 1.60 – 1.52 = 0.08 W
Competency Focused Practice Questions | Physics | Grade 12
35
[1 mark for the correct value of the current]
[1 mark for the correct value of the difference in the power]
[Note: Depending on number of decimal places taken, the answer to the sub‐
question (b) may vary between 1.52 and 1.54 and accordingly the answer to this
question may vary between 0.08 ‐ 0.06. Award marks accordingly.]
Q.42
(a) No, he will not be able to.
2
P = VI
Current drawn
I = P/V = 1200/220 = 5.45 A
Since the current drawn by steam iron is more than 5A, the circuit breaker will
trip.
[1 mark for the correct calculation of current]
[0.5 mark for the correct conclusion]
(b) The resistance of iron box:
P = V2/R
R = 220 x 220 /1200 = 40.3 ohm
[0.5 mark for the correct result]
Q.43
(a) BEFORE the second set of speakers were connected:
3
Total resistance = 10 + 10 = 20 Ω
Current drawn from the stereo output = I = 10/20 = 0.5 A
Since the two 10 Ω speakers are in series, the current through each one of them
is 0.5 A.
[1 mark for the correct value of current through the series combination]
[0.5 mark for the correct conclusion]
Competency Focused Practice Questions | Physics | Grade 12
36
(a) AFTER each of the two speakers of the second set was connected in parallel
to the first set of speakers:
The total voltage across the first set of speakers remains the same as earlier, that
is, 10 V.
So the current through each of the two 10 Ω speakers of the first set remains the
same as earlier.
That is, 0.5 A.
[0.5 mark for the correct result]
(b) No change in the loudness of the first set of speakers, as the current and the
power dissipated through each one of them remains unchanged even after the
second set of speakers are connected.
[0.5 mark for the correct conclusion]
(c) Any ONE of the following points:
‐ The amount of current drawn from the stereo output increases.
‐ Additional power is dissipated through the two speakers of the second set.
[0.5 mark for anyone correct point]
Q.44
(a) Resistance of the lamp, R = 2 + 2 = 4 ohm
2
Current through each lamp = I = V/R = 20/4 = 5 A.
[1 mark for the correct value of current]
(b) Energy consumed by the lamp in one second = VI = 20 x 5 = 100 J
4000 kJ of battery energy will last for 4000000/100 = 40000 s ~ 11 hours.
So the bulbs burn out before the battery drains out.
[1 mark for the correct result]
Competency Focused Practice Questions | Physics | Grade 12
37
Q.45
The maximum voltage that can appear across 16 ohms:
3
P = V12/R1
= 40 volt
[0.5 marks for correct calculation of maximum voltage that can appear across 16
ohms]
Equivalent resistance R2 across C and D:
R2 = 5/8 ohm
[1 mark for correct calculation of equivalent resistance between C and D]
As R1 and R2 are in series,
V1/R1 = V2/R2
= 40 x 5/(16 x 8)
= 25/16
= 1.56 volt
[1 mark for correct calculation of V2]
So the maximum voltage that can be applied across A and B,
V1 + V2 = 40 + 1.56 = 41.5 volt
[0.5 marks for correct final result]
Q.46
From the symmetry, the terminals C and D are at the same potential.
2
So the network can be reduced to the following:
[1 mark for the simplification point of the network]
Competency Focused Practice Questions | Physics | Grade 12
38
So 5 Ω, 6 Ω, 4 Ω, 4 Ω are in parallel.
Substituting and solving,
1/Req = 13/15
Req = 15/13
[1 mark for correct calculation of Req]
Q.47
For balanced condition,
2
[1 mark for the calculation of the correct value of req]
Solving for n,
n = 1/2
[1 mark for the calculation of the correct value of n]
Q.48
i. For VB – VD = 8 V:
4
Let i be the current through the loop in the anticlockwise direction,
Applying Voltage rule along the arm BAD
‐3i – 5 = ‐ 8
So i = 1 A
[1 mark for correct application of the voltage rule]
Applying Voltage rule along the arm BCD
‐10 + iR = ‐8
‐10 + R = ‐8
Competency Focused Practice Questions | Physics | Grade 12
39
so R = 2 Ω
[1 mark for correct application of the voltage rule]
ii. For VA ‐ VC = 8 V :
Let i' be the current through the loop in the anticlockwise direction,
Applying Voltage rule along the arm ABC
3i' – 10 = ‐ 8
So i' = 2/3 A
[1 mark for correct application of the voltage rule]
Applying Voltage rule along the arm ADC
‐5 ‐ i'R = ‐8
‐5 ‐ 2R/3 = ‐8
so R = 9/2 Ω
[1 mark for correct application of the voltage rule]
Competency Focused Practice Questions | Physics | Grade 12
40
3. MAGNETIC EFFECTS OF CURRENT AND MAGNETISM
Q. No
Question
Marks
Multiple Choice Question
Q.49
Read the following points of comparison between electric and magnetic fields
carefully.
1
i. Both Coulomb’s law and Biot‐ Savart’s laws follow inverse square law.
ii. Both electric and magnetic field values depend upon the angle between the
displacement vector and the field source.
iii. Electric field is produced by a scalar source whereas magnetic field is
produced by a vector source.
iv. The direction of the magnetic field is along the displacement vector joining
the field source and field point. On the other hand, the electric field is
perpendicular to the plane containing the displacement vector joining the field
source and the field point.
Identify the correct statements.
Q.50
A.
only i and ii
B.
only iii and iv
C.
only ii and iv
D.
only i and iii
A deuteron and an alpha particle move with the same kinetic energy under the
effect of identical magnetic fields.
1
What will be the ratio of the radii of their paths followed?
Q.51
A.
1
B.
√2
C.
1/2
D.
2
A particle of charge q and mass m projected with a speed v into a uniform
magnetic field B undergoes circular motion under the effect of magnetic force.
1
Identify an INCORRECT statement.
Competency Focused Practice Questions | Physics | Grade 12
41
A. Both KE and PE of this charged particle remain constant. So no work is done
by the magnetic field on it.
Q.52
B.
An identical charged particle projected at an angle other than 90o with
magnetic field B will undergo helical motion.
C.
Another charged particle of a different specific charge, projected with the
same speed will still have the same angular frequency.
D.
An identical charged particle projected with double the speed in the same
magnetic field will follow a circular path of larger circumference than the
earlier one but with the same time period.
A negatively charged particle moves with velocity v through a magnetic field B
and experiences a magnetic force F.
1
In each of the diagrams below, the representation of magnetic field lines is
missing.
Identify the probable direction of magnetic field lines in each case as Left, Right,
Upwards or Downwards.
Q.53
A.
i‐Right, ii‐Upwards, iii‐Upwards, iv‐Right
B.
i‐Left, ii‐Downwards, iii‐Downwards, iv‐Left
C.
i‐Downwards, ii‐Right, iii‐Upwards, iv‐Upwards
D.
i‐Left, ii‐Right, iii‐Upwards, iv‐ Downwards
Each of the diagrams represents magnetic field lines around a wire
1
carrying current I from different perspectives.
Identify which of the following representations are correct.
Competency Focused Practice Questions | Physics | Grade 12
42
Q.54
A.
only i and iv
B.
only ii and iii
C.
only i and iii
D.
only ii and iv
A rod of insulating material of length 0.5 m carries a static charge of 1 C at one
end. The rod is rotated in a vertical plane about a horizontal axis passing through
the other end with an angular frequency of 10π rad/s.
1
What is the magnetic field at the center of the circular path?
Q.55
A.
2π x 10‐6 T
B.
4π2 x 10‐6 T
C.
0.4 x 10‐7 T
D.
Zero
A wire of length L carrying a current I can be turned into a circular coil of N turns.
For what turned value of N, will the magnetic moment of this current carrying
coil be maximum?
A.
one
B.
4πL
C.
infinite
D.
(Magnetic moment is a constant for a given L and is independent of N)
Competency Focused Practice Questions | Physics | Grade 12
1
43
Q.56
Q.57
Which of the following statements is INCORRECT for a magnetic dipole?
A.
A freely suspended magnetic dipole in a uniform magnetic field always
turns to align parallel to the external magnetic field.
B.
A magnetic dipole that is free to move in a non‐uniform magnetic field can
slide as well as rotate.
C.
The torque required to hold a freely suspended magnetic dipole is
maximum when the dipole is perpendicular to the direction of the external
magnetic field.
D.
A small angular displacement given to a magnetic dipole result in a simple
harmonic motion about its original position, irrespective of its orientation
in a uniform magnetic field.
Two blocks of different materials are placed in a uniform magnetic field B. The
magnetic field lines passing through the two blocks are represented as follows.
1
1
Identify the suitable values of relative permeability μr and magnetic
susceptibility χ for the materials I and II.
Q.58
A.
For I : μr > 1, χ < 0, For II : μr < 1, χ > 0
B.
For I : μr < 1, χ = 0, For II : μr > 1, χ = 0
C.
For I : μr = 0, χ = 1, For II : μr = 1, χ = 0
D.
For I : μr < 1, χ < 0, For II : μr > 1, χ > 0
An electromagnet is a current‐carrying solenoid with a core as a magnetic
material. Given below are B‐H (net field Vs. magnetizing field) curves of three
different magnetic materials.
Competency Focused Practice Questions | Physics | Grade 12
1
44
Identify the most suitable magnetic material using their B‐H curves, that can be
used for the purpose of an electromagnet core.
Q.59
A.
I is better than II and III
B.
II is better than I and III
C.
III is better than I and II
D.
II and III are better than I
Read the following statements carefully:
1
I. Steel has high retentivity, high coercivity, and high permeability.
II. Soft iron has higher permeability, lower retentivity, and lower hysteresis loss
compared to steel.
III. Amorphous metals are non‐crystalline solids of very high resistivity and allow
very less hysteresis losses (example: an alloy of Fe, Ni, Co and glass). Their B‐H
(net field Vs. magnetizing field) curve is very narrow.
Based on the above material‐specific properties, identify which materials are
MOST suitable for the given electrical applications.
Competency Focused Practice Questions | Physics | Grade 12
45
I. Steel
A. Transformer core
II. Soft iron
B. Permanent magnets
III. Amorphous metals
C. Electromagnet
A.
I – A, B
II – B, C
III – C, A
B.
I–B
II – C
I–B
II – C, A
I–A
II – C
III – A
C.
III – A, B
D.
III – B
E.
Q.60
Consider the standard values of magnetic properties of three unknown metals.
Metal P
μr = 1.0003
χ = 3 x 10‐4 SI units
Metal Q
μr = 0.99
χ = ‐ 9.63 SI units
Metal R
μr = 5000
χ = 4.9 x 103 SI units
1
Identify which of the following does NOT suggest the expected behaviour of
these metals.
A.
With the rise in temperature, the μr value of R changes whereas for Q, it
remains constant.
B.
When placed in an identical external magnetizing field, the lines of
magnetic field through metal R become denser as compared to metal P.
C.
When placed in identical external magnetizing field B, the net magnetic
field inside Q is more than B whereas the net magnetic field inside P is less
than B.
D.
If the applied external magnetizing field B is non‐uniform and the metals
are free to move, Q moves from stronger to weaker field points whereas R
moves from weaker to stronger field points.
Competency Focused Practice Questions | Physics | Grade 12
46
Q.61
A freely suspended magnetic dipole is aligned along the magnetic meridian. Four
points A, B, C and D are located at equal distances from the centre of the
magnetic dipole towards the north, east, south, and west of the dipole
respectively.
i.
Direction of the magnetic field due to the dipole is same at A and C.
ii.
Direction of the magnetic field due to the dipole is same at A and D.
iii.
Direction of the magnetic field due to the dipole is opposite at B and D.
iv.
Direction of the magnetic field due to the dipole is opposite at C and D.
v.
Direction of the magnetic field due to the dipole is same at B and D.
vi.
Direction of the magnetic field due to the dipole is opposite at A and C.
1
Select the correct option.
Q.62
A.
Statements (iii) & (vi) are correct
B.
Statements (iv) & (v) are correct
C.
Statements (i), (ii) & (iii) are correct
D.
Statements (i), (iv) & (v) are correct
When a ferromagnetic material goes through the hysteresis loop of
magnetization, which of the following changes in magnetic susceptibility value
is/are possible?
(i)
1
Magnetic susceptibility of a ferromagnetic material is independent of
magnetizing field.
(ii) Magnetic susceptibility of a ferromagnetic material is zero through the
entire loop.
(iii) Magnetic susceptibility of a ferromagnetic material is infinite through the
entire loop.
(iv) Magnetic susceptibility of a ferromagnetic material is negative through the
entire loop.
(v) Magnetic susceptibility of a ferromagnetic material can be zero, negative,
positive, or infinity at different parts of the loop.
A.
Only (i) is correct.
B.
Only (ii) is correct.
C.
Only (v) is correct.
D. Either (ii) or (iii) or (iv) could be correct depending upon the material.
Competency Focused Practice Questions | Physics | Grade 12
47
Free Response Question/Subjective Question
Q.63
Two identical coils with a common centre are oriented at an angle of 45o to each
other. Both the coils have the same radius 'a', the same number of turns 'n', and
carry the same current 'I'.
2
What is the value of the resultant magnetic field at their center O?
Q.83
5
Three particles are projected into identical magnetic fields.
Particle
Proton
Deuteron
Alpha
particle
Speed of projection
106 m/s
106 m/s
106 m/s
Angle between v and B
30o
90o
60o
Charge q
e
e
2e
Mass M
m
2m
4m
Answer the following. Show the working in each case.
Q.65
i.
Which of the particles will revolve along the circular paths with maximum
frequency?
ii.
Identify the particle/s that will follow a spiral path.
iii.
Which particle/s revolves around the curved path of maximum radius?
Two parallel current‐carrying wires are placed at a perpendicular distance of r
from each other.
a.
Find the net magnetic field at the midpoint O if current I1 > I2.
b.
What is the effect on the net magnetic field at O in case if current I1 < I2?
Competency Focused Practice Questions | Physics | Grade 12
3
48
Q.66
A straight wire of length 4 m carrying a current of 0.5 A can be turned into either
a square or a circular loop of 2 turns, before placing it in a magnetic field of
intensity 0.1 T.
2
Which loop do you think will require less counter torque in order to hold it in a
position such that the axis of the loop is perpendicular to the magnetic field?
Find the value of this counter‐torque.
Q.67
A stream of singly charged particles of mass m1 = 0.8 x 10‐26 kg accelerated
through a potential difference V are projected into a uniform magnetic field B1
= 0.2 T. The stream deflects along a curved path under the effect of the magnetic
field and strikes the detector.
Competency Focused Practice Questions | Physics | Grade 12
3
49
Another stream of singly charged particles of mass m2 = 0.2 x 10‐26 kg, projected
through the same accelerating potential and into the same magnetic field B1, fail
to reach the detector.
To what value should the magnetic field be changed so that this stream of
particles strikes the detector?
Q.68
A wire carrying a current I = 200 A in the upward direction is placed in a magnetic
field B = 2 x 10‐3 T as shown in the figure.
3
A null point represents the location of a point with zero net magnetic field.
(a) On which side of the wire will a null point be located?
(b) On which side of the wire will a null point be located if the direction of applied
magnetic field B is reversed? Calculate the perpendicular distance of the null
point from the wire.
Competency Focused Practice Questions | Physics | Grade 12
50
Q.69
An electric field E = 2000 N/C is directed vertically downwards and a horizontal
magnetic field B is directed eastwards in a given velocity selector.
2
A stream of charged particles projected southwards at a speed of 2 x 103 m/s is
able to pass through this velocity selector completely undeflected.
A different particle of charge +3 μC, when projected southwards with a speed v
in the above magnetic field, experiences a net force of 2 x 10‐3 N, pointing
vertically downwards.
Find v.
Q.70
A uniform magnetic field B, directed along negative z‐axis and of value 2 x 10‐3
T extends through 1 meter along x‐axis, beyond which it becomes non‐uniform.
A singly charged ion of mass 1.6 x 10‐26 kg is projected along the positive x‐axis
into the magnetic field.
2
What should be the maximum velocity v of projection of the ion such that it just
fails to enter the non‐uniform part of the magnetic field region, but instead
emerges out of the uniform magnetic field region with a velocity v along the
negative x‐axis?
Q.71
A loop of area A carrying a current I when placed in an external magnetic field B
experiences torque.
4
(a) What is the direction of the torque with respect to the plane of the loop?
(b) Can the loop rotate around itself like a spin wheel due to this torque? Give a
reason for the answer.
(c) What is the shape of the graph between torque and angle θ, as it varies
between 0 and 180o? Represent it pictorially.
(d) State the orientation of the current‐carrying loop with respect to the external
magnetic field in which the total magnetic flux through its area is minimum.
Does this orientation constitute the stable or unstable equilibrium of the loop?
Q.72
A long straight current‐carrying wire is placed in proximity to a current‐carrying
solenoid of 100 turns per unit length.
4
(a) Identify suitable orientation of the current‐carrying solenoid with respect to
the current‐carrying wire, such that the net magnetic field along the axis of the
solenoid is zero. State the suitable directions of the current in each one also.
(b) If the ratio of the currents through the straight wire and solenoid is 200, find
the perpendicular distance between the straight current‐carrying wire and the
axis of the solenoid, which ensures a net zero magnetic field along the axis of
the solenoid.
Competency Focused Practice Questions | Physics | Grade 12
51
Q.73
A uniform magnetic field B = 0.002 T acts on a 2 cm long section PQ of an
insulated wire. The wire is attached to a spring of spring constant 0.8 N/m as
shown in the figure.
3
(a) What value of current should flow through PQ such that the spring is
stretched by 2 x 10‐4 m?
(b) Identify the possible direction of current through section PQ.
Q.74
A rotating table of diameter 0.5 m has a wire stretched across its surface passing
through its center. An electric current of 0.25 A flows through this wire and a
magnetic field of 0.8 T is applied in the plane of the rotating table.
2
(a) Does the magnetic force acting on the wire depend on the angle between
the wire and magnetic field?
(b) Find an angle between the wire and magnetic field such that the magnetic
force on the wire is equal to 0.05N.
(c) What is the effect on the magnetic force in case the magnetic field is applied
perpendicular to the plane of the table? Give a reason for your answer.
Q.75
P and Q are two identical charged particles of mass 4 x 10‐26 kg and charge 4.8x
10‐19 C, each moving with the same speed of 2.4 x 105 m/s as shown in the
figure. The two particles are equidistant from the vertical y‐axis. At some
instant, a magnetic field B is switched on so that the two particles undergo head‐
on collision.
Competency Focused Practice Questions | Physics | Grade 12
2
52
Find (a) the direction of the magnetic field and (b) the magnitude of the
magnetic field applied in the region.
Q.76 Two magnets of magnetic moment m1 = 1 Am2 and m2 = 4 Am2 respectively
are directed along the magnetic meridian as shown.
3
If the point P, along the magnetic meridian at a distance d1 = 20 cm from the
center of magnet 1, must be maintained at a zero magnetic field intensity due
to magnets 1 and 2, then,
(a) Find the suitable perpendicular distance d2 at which magnet 2 should be
placed without disturbing magnet 1. Take the horizontal component due to
Earth’s magnetic field as BH = 0.3 x 10‐4T.
The final answer may be expressed in fraction.
(b) Describe the consequence of reversing the polarities of the magnet 2 on
point P and the distance d2 from magnet 2.
Q.77 A magnetic dipole of dipole moment m is aligned parallel to an external
magnetic field B. Work of 0.25 J has to be done in order to turn it through angle
of 60o.
2
Find the external counter torque that is required in order to maintain the dipole
at this angle.
Competency Focused Practice Questions | Physics | Grade 12
53
Q.78 A magnet of magnetic moment m is tied to another magnet of moment 3m, with
similar poles above each other. The combination is freely suspended by a string
in a steady magnetic field and upon disturbance oscillates with a time period of
5 s.
3
Now if the magnet of moment 3m has its polarities reversed in the above
combination, determine the time period of the oscillation.
Q.79 At place A, where the angle of dip is 45o, a magnet oscillates with a frequency
of 12 oscillations per minute and at place B, where the angle of dip is 60o, the
magnet oscillates with a frequency of 10 oscillations per minute. If the total
magnetic field strength of Earth’s field at A is 0.3 x 10‐4 T, find its value at place
B.
2
Q.80 Given below are the hysteresis loops or B‐H (net magnetic field Vs. magnetizing
field) curves of two magnetizing materials.
3
(a) How do the residual magnetic field in (A) and (B) compare when the external
magnetizing field H is reduced to zero?
(b) In which specimen is the magnetic field required to destroy the residual
magnetism higher?
(c) Which of the two materials will undergo a greater loss of energy for every
cycle of magnetization? Give a reason for your answer.
(d) Which of the two materials is suitable for the making of electromagnets?
Give a reason for your choice.
Competency Focused Practice Questions | Physics | Grade 12
54
Q.81 A freely suspended bar magnet aligned in the magnetic meridian oscillates in
the horizontal plane with a time period of 0.2 s. A vertical current‐carrying wire
is placed at a perpendicular distance of 10 cm from the centre of the magnet.
3
If a current of 6 A flows through the wire upwards, what is the new time period
of oscillation of the bar magnet? Take horizontal component of Earth’s magnetic
field = 24 μT.
Q.82 The magnetic field inside a current‐carrying toroid is Bo. When the space inside
the toroid is filled with a material of susceptibility χ, the magnetic field along its
core becomes B.
2
Show that the percent increase in the magnetic field inside the toroid is 100
times χ.
Competency Focused Practice Questions | Physics | Grade 12
55
Q.83 An alpha particle is moving with a velocity v. It enters a magnetic field (B) as
shown below. The magnetic field is perpendicular and into the plane of paper.
2
.
A uniform electric field is applied in the same region as the magnetic field so that
the alpha particle passes undeviated through the combined fields.
(a)
What should be the direction of the electric field?
(b)
Without any change in the electric and magnetic field, the alpha particle is
replaced by the following particles:
(i)
proton moving with a velocity v
(ii)
electron moving with a velocity v/2
Will there be any change a deviation in the path of the particles? Give a reason
for your answer.
Competency Focused Practice Questions | Physics | Grade 12
56
Answer Key & Answer Scheme
Q. No
Answers
Marks
Q.49
D. only i and iii
1
Q.50
B. √2
1
Q.51
C. Another charged particle of a different specific charge, projected with the
same speed will still have the same angular frequency.
1
Q.52
B. i‐Left, ii‐Downwards, iii‐Downwards, iv‐Left
1
Q.53
A. only i and iv
1
Q.54
A. 2π x 10‐6 T
1
Q.55
A. one
1
Q.56
D. A small angular displacement given to a magnetic dipole results in a simple
harmonic motion about its original position, irrespective of its orientation in a
uniform magnetic field.
1
Q.57
D. For I : μr < 1, χ < 0, For II : μr > 1, χ > 0
1
Q.58
A. I is better than II and III
1
Q.59
B.
1
I
–
B
II
–
C
III
–A
Q.60
C. When placed in identical external magnetizing field B, the net magnetic field
inside Q is more than B whereas the net magnetic field inside P is less than B.
1
Q.61
D. Statements (i), (iv) & (v) are correct
1
Q.62
C. Only (v) is correct.
1
Competency Focused Practice Questions | Physics | Grade 12
57
Q.63
The magnetic field at the center O of coil 1 :
2
The magnetic field at the center O of coil 2:
[0.5 mark for each correct formula]
The resultant magnetic field at the center O:
Substituting and taking angle θ = 45,
[1 mark for correct calculation and final result]
Q.64
i. Frequency of revolution
5
Since B is the same for all,
Frequency n is proportional to q/m
np proportional to e/m
nd proportional to e/2m
nα proportional to 2e/4m, that is, e/2m
Proton will have a maximum frequency of revolution.
(1 mark for the correct conclusion)
ii. The particle that has velocity component parallel to magnetic field direction
will have a spiral path.
[0.5 mark for a correct statement of the condition for a particle to move along
spiral path]
Proton is projected at 30o to B: The velocity component parallel to B, that is,
v cos30 = √3.v/2. So proton will follow a spiral path.
The deuteron is projected at 90o to B: The velocity component parallel to B, that
is, v cos90 = 0. So deuteron will follow a circular path.
Competency Focused Practice Questions | Physics | Grade 12
58
Alpha particle is projected at 60o to B: The velocity component parallel to B, that
is, v cos60 = v/2. So the alpha particle will follow a spiral path.
[0.5 mark for each conclusion]
iii. Radius of the circular path =
Since B is same for all particle,
For proton,
For deuteron,
For alpha particle,
Deuteron has the maximum radius.
[0.5 mark for each of the correct statement of radius]
[0.5 mark for the correct conclusion]
Q.65
a. B1 at O due to I1 is
3
acting into the plane of the paper.
B2 at O due to I2 is
acting out of the plane of the paper.
[0.5 mark each for correct formula and the direction of the magnetic field due
to each current‐carrying wire]
If I1 > I2, then B1 > B2 and net magnetic field,B
= B1 – B2 =
Competency Focused Practice Questions | Physics | Grade 12
59
acting into the plane of the paper.
[1 mark for the correct formula of the net magnetic field along with the
direction]
b. If I1 < I2, then B2 > B1 and net magnetic field changes its direction.B
= B2 – B1 =
acting out of the plane of the paper.
[1 mark for the correct formula and direction of net magnetic field]
Q.66
Length of wire = 4 m
2
The perimeter of the coil with 2 turns = 2 m
For a given perimeter, a circular loop will have more area than the square loop.
Torque on the loop is directly proportional to the area of the loop.
Therefore, the counter‐torque required to hold the coil in a position such that
the axis of the loop is perpendicular to the magnetic field will be less for square
loop than for the circular loop.
[1 mark for the correct conclusion of lesser counter‐torque with correct
argument]
The counter torque required is
τ = MB sin 90 = n I A B = 2 x 0.5 x (side x side) x 0.1
Side of square = perimeter /4 = 2/4 = 0.5 m
τ = 2 x 0.5 x (0.5 x 0.5) x 0.1
= 0.0250 Nm.
[1 mark for the correct calculation of the counter torque]
Q.67
Equating the kinetic energy of charged particles to the energy gained due to
accelerating potential V,
Competency Focused Practice Questions | Physics | Grade 12
3
60
[1 mark for the correct expression of speed]
Equating the magnetic force on the charged particles to the centripetal force
acting on them,
[1 mark for the correct expression of magnetic field]
For same accelerating potential V, radius r and charge q,
B2 = B1/2 = 0.2/2 = 0.1 T
[1 mark for correct calculation of value of B2]
Q.68
(a) On the left side of the wire.
3
[0.5 mark for the correct location of null point]
(b) On the right side of the wire.
[0.5 mark for the correct location of null point]
Null point is the point where the magnetic field due to the current‐carrying wire
is equal to the external magnetic field and is opposite in direction.
r = 0.02 m
[1 mark for correct statement of condition and the formula]
[1 mark for correct calculation and final result]
Q.69
In the first case, for the charged particles to go undeviated,
2
qE = qvB
So B = E/v = 2000/2 x 103 = 1 T
[1 mark for correct calculation of magnetic field B]
In the second case,
Net force, FN = FE ‐ FB = q (E ‐ vB)
Competency Focused Practice Questions | Physics | Grade 12
61
Substituting,
v = 4000/3 = 1.3 x 103 m/s
[1 mark for the correct calculation and final result]
Q.70
In order that the ion just fails to enter the non‐uniform part of the magnetic field
region, it should follow a circular path of radius, not more than 1 m.
2
So the radius of the path followed by the ion in magnetic field B,
[1 mark for the correct statement of the condition and the formula]
Substituting,
Solving for v,
v ≤ 2 x 104 m/s
[1 mark for correct calculation and final result]
Q.71
(a) Torque on the current‐carrying loop in the magnetic field always acts in the
plane of the loop.
4
[0.5 mark for the correct direction of torque]
(b) No.
For the loop to rotate around itself, it requires a torque that is along its vertical
axis.
But in this case, the torque is along the plane of the loop. So the loop cannot
rotate like a spinwheel.
[0.5 mark for the correct answer]
[1 mark for the correct reason]
(c) It is sinusoidal.
Competency Focused Practice Questions | Physics | Grade 12
62
[0.5 mark for correct shape]
[0.5 mark for the correct representation]
(d) The magnetic flux is minimum when the area vector of the current‐carrying
loop is antiparallel to the magnetic field.
This orientation constitutes unstable equilibrium.
[0.5 mark for the correct orientation]
[0.5 mark for the identification of this orientation as unstable]
Q.72
(a) Consider a straight current‐carrying wire in the plane of paper with the
current in the upwards direction.
4
Place the solenoid on the right side of the wire, such that its axis is perpendicular
to the plane of the paper.
The current in the solenoid should flow in an anticlockwise direction when
looked at from above.
In the above orientation, at a specific perpendicular distance between them, the
net magnetic field along the axis of the solenoid can be zero.
[1 mark for the correct description of the orientation and 1 mark for the correct
directions of the current]
OR
Alternatively.,
Consider a straight current‐carrying wire in the plane of paper with the current
in the downwards direction.
Place the solenoid on the left side of the wire, such that its axis is perpendicular
to the plane of the paper.
Competency Focused Practice Questions | Physics | Grade 12
63
The current in the solenoid should flow in an anticlockwise direction when
looked at from above.
In the above orientation, at a specific perpendicular distance between them, the
net magnetic field along the axis of the solenoid can be zero.
[1 mark for the correct description of the orientation and 1 mark for the correct
directions of the current]
(b) If the perpendicular distance of separation between the wire and axis of the
solenoid is r,
B due to current‐carrying wire,
B due to current carrying solenoid along its axis = μonIs
[0.5 mark for each correct formula]
For net magnetic field along the axis to be zero,
Substituting, Iw / Is = 200, n = 100
r = (1/π) m
[1 mark for the correct calculation and final result]
Q.73
(a) When the spring is stretched,
3
the restoring force on the spring upwards= the magnetic force acting on the wire
PQ downwards
[1 mark for the correct equation of the spring force with magnetic force]
kx = BIL
I = kx/BL = 0.8 x 2 x 10‐4/(0.002 x 0.02) = 4 A
[1 mark for the correct calculation and final result]
(b) Direction of current is from Q to P
[1 mark for the correct direction of the current]
Q.74
(a) Magnetic force on the wire is dependent on the angle between wire and
magnetic field.
2
[0.5 mark]
Competency Focused Practice Questions | Physics | Grade 12
64
(b) Force F = B I L sinθ
θ = 30o
[1 mark for calculation and correct final result]
(c) The value of magnetic force on the wire will remain constant as irrespective
of the rotation of the table, the angle between the direction of the current and
the magnetic field will always be 90o.
[0.5 mark for correct explanation]
Q.75
(a) Perpendicular and into the page.
2
[0.5 mark]
(b) For a head‐on collision to take place, the radius of the path of each ion should
be equal to 0.5 m.
Solving for B = 0.04 T
[0.5 mark for determining the value of r]
[1 mark for correct calculations & result]
Q.76
(a) Magnetic field B1 due to m1 at point P,
3
= 10‐7 x 2 x 1/(0.2)3 = 0.25 x 10‐4 T
Net magnetic field along horizontal direction at P (due to m1 and BH),
B1 + BH = 0.55 x 10‐4 T
[0.5 mark for expression of net field at P due to B1 and BH]
Magnetic field B2 due to m2 at point P,
= 10‐7 x 4/(d2)3
Competency Focused Practice Questions | Physics | Grade 12
65
[0.5 mark for expression of B2]
For net field intensity at P to be zero,
10‐7 x 4/(d2)3 = 0.55 x 10‐4 T
d2 = 0.193m = 19.3 cm
[1 mark for final equation and calculation]
(b) If the polarities of magnet 2 are reversed, the magnetic field B2 due to
magnet 2 will point in the same direction as the resultant of BH and B1. Hence
the magnetic field at point P will no longer remains zero, irrespective of the
distance d2.
Q.77 Work done to rotate from parallel to 60o,
2
W = | mB (cosθ2 ‐ cosθ1) |
= | mB (cos 60 – cos 0) |
= | mB (1/2 – 1) |
= mB/2
[1 mark for correct formula and final expression of W]
Counter torque required to hold the dipole at angle 60o ,
τ = mBsinθ = mB sin60 = √3mB/2
τ = √3 x W = √3 x 0.25 = √3/4 N‐m
[1 mark for correct calculation and final result]
Q.78 In the first arrangement, the net magnetic moment of the combination is
3
m1 = m + 3m = 4m
[0.5 mark for net magnetic moment]
The initial time period of oscillation,
[0.5 mark for the correct formula of T1]
The net magnetic moment of the combination with reversed polarities is m2 =
3m ‐ m = 2m
Competency Focused Practice Questions | Physics | Grade 12
66
[0.5 mark for net magnetic moment]
The time period of oscillation after reversing the polarities,
[0.5 mark for correct formula of T2]
T2 = 5√2 s
[1 mark for correct final result]
Q.79 The frequency of oscillation by a magnetic dipole in presence of external
magnetic field B is
2
here B = B1cos45 where B1 = 0.3 x 10‐4 T and f1 =12/60
here B = B2cos60 and f2 = 10/60
[0.5 mark each for correct formula and denoting correct values of magnetic field
B]
So
= 0.294 x 10‐4 T
[1 mark for correct calculation and final result]
Q.80 (a) When H = 0, the residual field in (A) is OQ.
3
When H = 0, the residual field in (B) is OQ’
Competency Focused Practice Questions | Physics | Grade 12
67
Hence, residual field in (A) < residual field in (B)
[0.5 mark for correct comparison]
(b) Specimen B
[0.5 mark for correct answer]
(c) Material (B) undergoes greater hysteresis loss than (A).
Hysteresis loss is energy dissipated per cycle per unit volume of the specimen
and is given by the area enclosed by the B‐H loop.
[0.5 mark for correct identification of the material]
[0.5 mark for a correct reason]
(d) Material (A) is suitable for the making of electromagnets.
Material (A) has lower retentivity as compared to material (B).
[0.5 mark for correct identification of the material]
[0.5 mark for a correct reason]
Q.81 Magnetic field due to a current‐carrying wire at a distance r,
3
B = (μo i)/2πr = 2 x 10‐7 x 6 / 0.1 = 12 x 10‐6 T = 12 μT
This field due to the current‐carrying wire acts from N to S, that is, opposite to
BH.
So the net magnetic field under which the bar magnet oscillates:
B’ = BH – B = 24 – 12 = 12 μT
[0.5 mark for the correct calculation of B due to current‐carrying wire]
[0.5 mark for correct net magnetic field acting on bar magnet]
The time period of oscillation of bar magnet in presence of a magnetic field,
[1 mark for the correct formula of time periods]
T2 = T1√2 = 0.2 x √2 = 0.2 x 1.414 = 0.28 s
[1 mark for correct calculation and final result]
Competency Focused Practice Questions | Physics | Grade 12
68
Q.82 A magnetic field without the core:
2
Bo = μoH
Magnetic field with the core:
B = μH = μo ( 1 + χ) H
[0.5 mark each for correct formula]
Change in a magnetic field,
B – Bo = μo χ H
% increase in a magnetic field,
Hence proved.
[1 mark for the correct final result]
Q.83 (a) downwards in the plane of the paper (or) perpendicular to B and v,
downwards
2
(b)
(i) proton moving with a velocity v
No deviation (0.5 marks)
qE = qvB
Force does not depend on mass and the charge cancels out. So the proton will
also pass undeviated. (0.5 marks for correct explanation)
(ii) electron moving with a velocity v/2
The electron will deviate upwards. (0.5 marks)
Since velocity is halved, electric force > magnetic force. (0.5 marks for correct
explanation)
Competency Focused Practice Questions | Physics | Grade 12
69
4. ELECTROMAGNETIC INDUCTION AND ALTERNATING CURRENT (AC)
Q. No
Question
Marks
Multiple Choice Question
Q.84 A current‐carrying coil C1 is paired with another coil. An emf ε is induced in coil
C2 when the current through C1 drops to zero from a non‐zero value within 1 s.
1
If the coil C1 is paired with C3, an emf 0.7ε is induced in it, for the same rate of
drop in the current through C1 as earlier.
What is the ratio of mutual inductance M12 (for the pair of coils C1 and C2) to
that of mutual inductance M13 (for the pair of coils C1 and C3)?
A.
0.7
B.
1
C.
1.42
D.
7
Q.85 Coil C carries a steady current. A second coil is placed in close proximity to coil
C in different configurations as shown.
1
Which of the following options represents the correct order of the mutual
inductance values for the pair of coils in given configurations?
Competency Focused Practice Questions | Physics | Grade 12
70
A.
(i) > (iii) > (iv) > (ii)
B.
(iv) > (iii) > (i) > (ii)
C.
(ii) > (iv) > (iii) > (i)
D.
(iii) > (i) > (ii) > (iv)
Q.86 A long solenoid S has length l1, area of cross‐section A and number of turns N1.
A coil C of shorter length l2, number of turns N2 but same area of cross‐section
is wound over solenoid S. If a current I flows through the coil C, then which of
the following represents the flux linked with the solenoid S?
1
A.
B.
C.
D.
Q.87 There is a pair of concentric and coplanar conducting loops of radii R1 and R2
such that R2 = 0.01 R1.
1
To which of the following is the mutual inductance M for this pair directly
proportional?
A.
1/R12
B.
R12
C.
1/R1
D.
R1
Q.88 A small conducting metal ring falls vertically down with its plane parallel to y‐z
plane. During the fall, the ring crosses a small region of the uniform magnetic
field directed along the x‐axis between the times t=0 and t = τ seconds.
1
Which of the following graphs represent the variation of current induced in the
ring during the fall?
Competency Focused Practice Questions | Physics | Grade 12
71
A.
B.
C.
Q.89
A capacitor is short‐circuited as shown.
1
Match the possible motions listed in [A] of the bar magnet relative to the
loop with the suitable responses listed in [B] due to the induced current,
if any.
Competency Focused Practice Questions | Physics | Grade 12
72
[A]
I. Magnet with N facing the loop moves towards the coil
II. Magnet with N facing the loop moves away from the coil
III. Magnet NS is stationary
IV. Magnet with S facing the loop moves towards the coil
V. Magnet with S facing the loop moves away from the coil
[B]
i. Excess electrons accumulate on plate a
ii. Excess electrons accumulate on plate b
iii. The plates a and b stay neutral
A. [I ‐ i] , [II ‐ ii ], [III ‐ iii], [IV ‐ i ], [V ‐ ii]
B. [I ‐ iii] , [II ‐ ii], [III – i], [IV – ii ], [V – i]
C. [I ‐ii] , [II ‐ i], [III ‐ iii], [IV ‐ i], [V ‐ii]
D. [I ‐ i] , [II ‐ ii], [III ‐ iii], [IV ‐ ii], [V ‐ i]
Q.90 A square loop of a single turn is placed with its plane perpendicular to a uniform
magnetic field. The magnetic flux through the square loop is 0.002 Wb.
1
The square loop is now reshaped into a circular loop. What is the flux (in Wb)
through the circular loop?
A.
π/4
B.
0.002
C.
0.008/π
D.
0.0005π
Q.91 If Pav represents average power dissipated and iav is the average current
through a resistor over one cycle of the input sinusoidal voltage, which of the
Competency Focused Practice Questions | Physics | Grade 12
1
73
following statements is correct for an ac circuit with resistor only?
A.
Pav = 0 & iav > 0
B.
Pav > 0 & iav = 0
C.
Pav > 0 & iav > 0
D.
Pav = 0 & iav = 0
Q.92 An LCR series circuit is connected to an ac supply of ω = 100 rad/s. Given the
values as R = 100‐ohm, L = 500 mH, C = 5 μF, study the following statements
carefully.
I. The given circuit (LCR) is dominantly capacitive
II. The instantaneous current in the circuit leads Vmax.
III. If w greater than 200√10 rad/s, the circuit becomes dominantly inductive
IV. The LCR circuit can be made capacitive or inductive by simply changing the
angular frequency of the input ac supply, keeping the voltage Vmax constant.
Identify the correct option.
Q.93
A.
Only statement IV is correct
B.
Only statements I and II are correct
C.
Only statements I and III are correct
D.
All statements are correct
In an ac circuit containing a resistor R, the power dissipated is P1.
1
If a capacitor C is added to the above circuit, the power dissipated through the
resistor changes to P2.
If an inductor L replaces the capacitor C, the power dissipated through the
resistor changes to P3.
If the capacitive reactance offered by the capacitor equals the inductive
reactance offered by the inductor, then which of the following represents the
correct relation between P1, P2 and P3?
A.
P1 = P2 = P3
B.
P3 > P1 > P2
C.
P1< [P2 = P3]
D.
P1 > [P2=P3]
Competency Focused Practice Questions | Physics | Grade 12
74
Q.94 A 5‐ohm resistor, a 5 mH inductor and a 5 μF capacitor, joined in series resonate
with an ac source of frequency ωo. If only the resistance is changed to 10 ohm,
the circuit resonates at a frequency ω1. If only the inductor is changed to 20 mH,
the circuit resonates at a frequency ω2.
1
Find the ratio ω1/ω2.
A. 0.5
Q.95
B.
1
C.
2
D.
4
Which of the ac circuits with the following input voltage and current dissipates
maximum power P?
A.
B.
C.
D.
Q.96
1
Input voltage Vo = 2 volt, Io = 4 ampere and phase angle Φ = π/4.
Input voltage V = Vo sinωt volt and the current I = Io sin(ωt ‐ π/2) ampere
Input voltage V = 2 cosωt volt and the current I = 4 sinωt ampere
Input voltage V = 100 sin100t volt and the current I = 100 sin(100t +π/3)
milliampere
The four graphs below represent instantaneous power dissipated across various
circuit elements such as resistor, capacitor & inductor connected either
individually or in a combination to an ac supply.
1
Study the following statements carefully and select the correct option. I.
Average power dissipated in (2) & (3) is zero.
II. Average power dissipated in (1) is that in a resistor and is given as Pav =
Irms.Vrms
III. Average power dissipated in (4) is that in an LCR combination circuit and is
always positive.
Competency Focused Practice Questions | Physics | Grade 12
75
IV. Graphs (1), (2) & (3) represent power dissipation in resistor, capacitor, and
inductor in an ac circuit respectively
A. Only I & II are correct
B. Only III & IV are correct
C. Only I, III & IV are correct
D. Only I, II & IV are correct
Q.97
In a circuit, a resistor of 1 kilo‐ohm is connected to an element X as shown. An
ac input voltage of Vi = 10 mV is applied at variable frequencies.
1
Given below is the data observed for output voltage Vo and total impedance Z
of the circuit as a function of input voltage frequency ν.
ν
Z
Vo
10 kHz
3 x 103 ohm
16.1 mV
100 kHz
1.04 x 103 ohm
1.6 mV
1 MHz
1.003 x 103 ohm
0.16 mV
10 MHz
103 ohm
16 mV
Study the following statements carefully and select the correct option given
below.
I.
The output current is expected to increase with the increase in input
frequency.
II.
The reactance offered by X increases with the increase in input frequency.
III. If the ac input is replaced by dc input of the same voltage, the output
voltage Vo becomes zero.
IV. The circuit element X could be either a capacitor or an inductor.
Competency Focused Practice Questions | Physics | Grade 12
76
A.
Only statements I & II are correct
B.
Only statements III & IV are correct
C.
Only statements I & III are correct
D.
Only statements II & IV are correct
Free Response Question/ Subjective Question
Q.98 A constant magnetic field B = 1 T acts perpendicularly on a static square copper
coil of 100 turns with each side of 10 cm. The coil is removed from the magnetic
field within 0.5 s and then replaced in another 0.5 s.
2
a. Calculate an average emf that develops in the coil during the entire motion
lasting for 1 s.
b. Find the net thermal energy dissipated in the coil during the entire motion.
Take resistance of the coil as 1 ohm.
Q.99 Given two coplanar and concentric loops 1 and 2 as shown.
2
A time‐varying voltage (3 + 2t) is applied to the larger loop 1. If the resistance of
the loops is R1 = 10 ohm and R2 = 1 ohm, then determine the current induced
in the smaller loop.
Q.100 Coil 1 has self‐inductance L1 which is 3 times the self‐inductance L2 of coil 2. If
during a certain instant, the rate of increase in current and the power dissipated
in these two coils is the same, then determine the ratio of
3
(a) their induced voltages
(b) currents
(c) energy stored in the two coils at that instant.
Competency Focused Practice Questions | Physics | Grade 12
77
Q.101 Sam wants to generate 10 mV of EMF by moving a wire at 5m/s through a steady
magnetic field region of 0.06 T.
1
a. What should be the angle between the magnetic field and direction of motion
so that the wire can be of the shortest length?
b. What should this shortest length of the wire be?
Q.102 The Earth’s magnetic field in the northern hemisphere has a downward
component that can be assumed to be uniform. Consider an airplane flying
through this magnetic field from India towards Russia in the north. A motional
emf develops across the tips of its two wings.
2
(a) Identify the polarities developed on the left‐wing and the right‐wing. Explain
the answer.
(b) An insulated wire along with a light bulb connects the tips of the right and
the left‐wing of the airplane. Will the bulb glow due to the induced emf? Explain
the answer.
Q.103 A projectile with an enclosed tiny magnet is projected with a speed v through a
pair of identical coaxial coils C1 and C2 as shown.
2
(a) Plot a graph of induced emf ∈ versus time t for the period of passage of the
projectile through two successive coils. Take the counter‐clockwise direction of
the induced current in a coil as seen from the starting point of the incoming
projectile as positive.
(b) If the time interval between two emf pulses observed in C1 and C2
respectively is 3 ms, determine the speed of the projectile.
Competency Focused Practice Questions | Physics | Grade 12
78
Q.104 A solenoid S of radius 50 cm with 100 turns per unit length is aligned along the
x‐axis carrying a current of 10 A.
2
A coil C of radius 10 cm is coaxially placed inside the solenoid such that it can
rotate about its diameter directed along the y‐axis. Refer to the diagram below.
If the coil C of 50 turns revolves with a constant angular speed of π rad/s,
determine the emf generated in coil C.
(Note: The final answer may be written in terms of constants μo and π)
Q.105 In a given LCR series ac circuit, the inductive and capacitive reactance are given
as 400 ohm and 600 ohm respectively. If resistance R in the circuit is 100 ohm
and the maximum input voltage of the ac supply is 100 V, then determine the
following:
5
(a) net impedance of the circuit.
(b) the maximum current in the circuit.
(c) the maximum voltages across each of the elements.
(d) the sum of the maximum voltages across the three elements.
(e) Is it possible to attain the value as obtained in part (d) in the given circuit?
Give a reason for your answer.
Q.106 A series LCR circuit is capacitive if the ac supply is at angular frequency ω1 with
the phase angle Φ = ‐30. The circuit becomes inductive if the ac supply is
changed to angular frequency ω2 while keeping the maximum value of source
voltage constant and the corresponding phase angle becomes Φ = 30.
3
What is the change in power delivered in the circuit due to a change in angular
frequency from ω1 to ω2? Explain your answer.
Competency Focused Practice Questions | Physics | Grade 12
79
Q.107 An LCR circuit is connected to an ac voltage of fixed Vmax and angular frequency.
Current through the circuit is Imax and the resistance R equals the inductive
reactance XL in the circuit.
2
Now if the distance of separation between the capacitor plates is doubled, the
current Imax in the circuit is reduced to half.
Determine the initial relation between resistance R and capacitive reactance Xc.
Q.108 An ac source with Vmax = 100 V and frequency 50 Hz is connected across a 1 μF
capacitor. Assuming that at time t = 0, the energy stored in the capacitor is zero,
determine the following:
2
(a) the current in the capacitor at time t = 1/100 s.
(b) maximum current in the circuit.
Q.109 The ratio of the turns in the primary and the secondary coil in a given
transformer is 1:10. If the load resistor draws a current of 1.2 A from the
secondary coil, determine the current in the primary coil, assuming that the
transformer has an efficiency of only 90%.
2
Q.110 A given ac power supply has a voltage rating of 80 V. An LCR series circuit draws
a current of 5A when connected to this ac power supply. Determine the range
of values of the average power that can be delivered by the ac power source to
the circuit.
2
Q.111 The household power supply is at 200 V and 60 Hz. Find the time required for:
2
a.
the value of current to change from 0 to its rms value
b.
the value of voltage to change from its rms value to 0.
Q.112 An alternating current is sent through a 100 km long telephone wire of
capacitance 0.01 μF/km at a frequency of 8 kHz. Find out what value of an
inductor that must be connected in series to this wire, so that the current
through the wire is maximum.
2
Q.113 Given is a coil of resistance R and inductance L connected to a power supply.
2
If the power supply is 50 V dc, a current of 0.5 A flows through the coil.
If the power supply is 50 V ac of frequency 50 Hz, a current of 0.2 A flows through
the coil.
Determine the resistance R and inductance L of the coil.
Competency Focused Practice Questions | Physics | Grade 12
80
Q.114 An inductor L is connected to an ac power source of frequency 'ν' Hz. The
2
inductor L undergoes alternate cycles of magnetization and demagnetization
due to sinusoidal variation of current through L.
If the time taken for a successive magnetization and demagnetization of the
inductor is 8 ms, determine the frequency 'ν' of the input ac power source.
Q.115 The rating of a filament bulb is given as 50W, 100V. This bulb is to be used in a
2
house where the power supply is at 200 V, 50 Hz.
Determine the inductance of the coil that must be connected in series to the
bulb, so that it glows to its maximum power rating, but doesn’t fuse.
Competency Focused Practice Questions | Physics | Grade 12
81
Answer Key & Marking Scheme
Q. No
Answers
Marks
Q.84 C. 1.42
1
Q.85 A.
1
(i) > (iii) > (iv) > (ii)
Q.86
1
C.
Q.87 D. R1
1
Q.88
1
A.
Q.89 C. [I ‐ii] , [II ‐ i], [III ‐ iii], [IV ‐ i], [V ‐ii]
1
Q.90 C. 0.008/π
1
Q.91 B. Pav > 0 & iav = 0
1
Q.92 D. All statements are correct
1
Q.93 D. P1 > [P2=P3]
1
Q.94 C. 2
1
Q.95 A. Input voltage Vo = 2 volt, Io = 4 ampere and phase angle Φ = π/4.
1
Q.96 D. Only I, II & IV are correct
1
Q.97 C. Only statements I & III are correct
1
Competency Focused Practice Questions | Physics | Grade 12
82
Q.98 a. Emf during removal
2
Φ1 = nBA = 100 x 1 x 0.1 x 0.1 = 1 Wb
Φ2 = 0
Time = 0.5 s
Induced emf |ε| = ΔΦ/Δt = 1/0.5 = 2 V
Emf during return path
Φ1 = 0
Φ2 = nBA = 100 x 1 x 0.1 x 0.1 = 1 Wb
Time = 0.5 s
Induced emf |ε| = ΔΦ/Δt = 1/0.5 = 2 V
Average emf during the entire motion
=0
Alternatively,
Φ1 = nBA = 100 x 1 x 0.1 x 0.1 = 1 Wb
Φ2 = nBA = 100 x 1 x 0.1 x 0.1 = 1 Wb
Change in flux = 0
Time = 1 s
Average induced emf during the entire motion = 0
[1 mark for the correct calculation of average induced emf]
b. Thermal energy dissipated
= i2rt = ε2t/r
Energy dissipated during withdrawal: 22 x 0.5/1 = 2 J
Energy dissipated during return path: 22 x 0.5/1 = 2 J
Total energy dissipated during the entire motion = 4 J
[1 mark for the correct calculation of total energy dissipated]
Competency Focused Practice Questions | Physics | Grade 12
83
Q.99
Current in loop 1
2
I1 = V/R1 = (3+2t)/10
Magnetic field at the center of the loop 1 due to current I1,
B1 = μoI1/2p = μo(3+2t)/20
[0.5 mark for correct result of magnetic field]
Flux linked with loop 2,
Emf induced in loop 2,
Induced current in loop 2,
[0.5 mark for each of the correct results]
Q.100 (a) As e = L dI/dt
3
e1/e2 = L1/L2 = 3L/L = 3
[1 mark for the correct ratio]
(b) As power P = eI
I1/I2 = P1e2/P2e1 = e2/e1 = 1/3
[1 mark for the correct ratio]
(c) Energy stored in a coil, U = ½ LI2
U1/U2 = ½ L1I12/½ L2I22 = 3(1/3)2 = 1/3
[1 mark for the correct ratio]
Q.101 a. The wire can be of the shortest length if it is moved perpendicular to the
magnetic field
1
[0.5 mark for the correct identification of the angle]
b. EMF = Blv
Competency Focused Practice Questions | Physics | Grade 12
84
. l = EMF / Bv
= 10 x 10‐3 / 0 x 5 = 1/30 m
= 3.3 cm
[0.5 mark for the correct calculation of the length]
Q.102 (a) As the plane flies northwards, the right‐hand rule indicates that the positive
charge experiences a force directed toward the west.
2
Thus, the left wingtip becomes positively charged and the right wingtip
negatively charged.
[1 mark for the correct identification of the polarities]
(b) The light bulb will not glow.
The wings with the connecting insulated wire and the bulb constitute a single‐
loop circuit.
As the plane flies through a uniform Earth’s magnetic field, the magnetic flux
through this loop is constant and net emf generated is zero.
[0.5 mark for the correct conclusion] [0.5 mark for the correct reason]
Q.103
2
(a)
[1 mark for the correct graph]
(b) Speed of the projectile
= d/t = 1/(3 x 10‐3)=333.3 m/s
[1 mark for the correct calculation of the speed of the projectile]
Competency Focused Practice Questions | Physics | Grade 12
85
Q.104 Magnetic field along axis of solenoid = B = μo x 100 x 10 = 1000μo
2
Magnetic flux through coil C,
Φ = NB.A = NBA cosωt
[0.5 mark each for magnetic field and flux]
Emf induced in the coil due to its rotation, ∈ = ‐dΦ/dt = NBAω sinωt
Upon substitution,
∈ = 500 μo π2 sinπt
[1 mark for correct induced emf]
Q.105 (a) Impedance
5
=100√5 ohm
[1 mark for correct result]
(b) imax = Vmax/Z = 100/100√5
= 1/√5 A
[1 mark for correct result]
(c) VR,max = imax x R = 100/√5 volt
VL,max = imax x XL = 400/√5 volt
VC,max = imax x XC = 600/√5 volt [0.5
mark for each correct result]
(d) The sum of VR,max + VL,max + VC,max = 100/√5 + 400/√5 + 600/√5 =
1100/√5 ∼ 493 V
[0.5 mark for correct result]
(e) The sum of voltages across the three elements (= 493 V) in LCR series circuit
exceeds the maximum input voltage (100 V) of the circuit. This is physically
impossible.
[0.5 mark for correct conclusion]
The voltages across each of the elements need to be added after considering
their respective phase angles. Each of the maximum values across the elements
L, C and R is attained at different instants of time. Hence the net voltage equal
to 493 V is never achievable at any given instant during any one ac cycle of
circuit.
[0.5 mark for correct reason statement]
Competency Focused Practice Questions | Physics | Grade 12
86
Q.106 Power dissipated in the circuit remains unchanged.
3
[0.5 mark for correct conclusion]
As Power, Pav = Irms. Vrms. cosΦ
Power factor cos Φ remains unchanged.
With the change in ω1 ‐>ω2
Φ ‐> ‐Φ
cosΦ = cos (‐Φ), so power factor is unchanged.
[1 mark for proving that cosΦ remains unchanged]
Also, Vrms remains unchanged as the source voltage is constant.
[0.5 mark for proving that Vrms remains unchanged]
Irms remains unchanged.
When Φ = ‐30
tan ‐30 = (XL ‐ XC)/R
XL ‐ XC = ‐R/√3
When Φ = 30
tan 30 = (XL ‐ XC)/R
XL ‐ XC = R/√3
Since impedance is
In both cases Z = 2R/√3.
Since Irms = Vrms/Z
So Irms also remains unchanged.
[1 mark for proving that Irms remains unchanged]
Hence the Power, Pav = Irms.Vrms.cosΦ remains unchanged.
Competency Focused Practice Questions | Physics | Grade 12
87
Q.107 For the distance between the plates doubled,
2
C’ = kA/d = C/2
So capacitive reactance X’c = 2XC
[1 mark for getting the correct relation between XC and XC']
For the current Imax to be reduced to half,
Impedance Z’ = 2Z
Substituting R = XL and solving, we get
XC = 3R/2
[1 mark for correct calculations and final result]
Q.108 (a) XC = 1/Cω
2
= 1/(10‐6 x 2π x 50) = 104/π ohm
[0.5 mark for correct formulae]
ic = (π/100). sin (π + π/2)
= ‐(π/100) sinπ/2
ic = ‐ π/100 A
[1 mark for correct calculations and result]
(b) imax = Vmax/XC = Vmax. ωC
= 100 x 2π x 50 x 10‐6 = π/100 A
[0.5 mark for correct result]
Competency Focused Practice Questions | Physics | Grade 12
88
Q.109 Efficiency = Output power / Input power
2
0.90 = (Vs Is)/(IpVp )
[1 mark for correct formulation]
Ip = 100 Vs Is /90.Vp
Substituting and solving,
Ip = 13.3 A
[1 mark for correct substitution and calculation]
Q.110 Average power drawn by LCR circuit = Pav = Vrms. Irms. cosΦ = 80 x 5 cosΦ =
400 cosΦ
2
[1 mark for correct representation of Pav]
In a series LCR circuit:
The maximum value of the power factor, cosΦ is 1, in the case of resonance
wherein Φ is 0.
The minimum value of the power factor, cosΦ approaches 0 as the value of Φ
approaches either ‐90 or 90 depending upon whether the circuit is inductive or
capacitive.
[0.5 mark for the correct reasoning of variation of the value of power factor in
an LCR circuit]
Pav values range between 0 and 400.
0 ≤ Pav ≤ 400
[0.5 mark for determination of the correct range of values]
Q.111 a. Current i = io sin2πνt
2
At t = 0, i = 0
At time = t, i = irms = io/√2
io/√2 = io sin2πνt
π/4 = 2πνt
Solving for t,
t = 1/480 = 2 x 10‐3 s
[0.5 mark for correct formula and 0.5 mark for correct result]
b. Time taken by current to change from 0 to rms value
Competency Focused Practice Questions | Physics | Grade 12
89
= Time taken by the voltage to change from 0 to rms value
= Time taken by voltage to change from rms value to 0.
So t = 2 x 10‐3 s as above.
[0.5 mark for correct reasoning & 0.5 mark for correct final result]
Q.111 c. Current i = io sin2πνt
2
At t = 0, i = 0
At time = t, i = irms = io/√2
io/√2 = io sin2πνt
π/4 = 2πνt
Solving for t,
t = 1/480 = 2 x 10‐3 s
[0.5 mark for correct formula and 0.5 mark for correct result]
d. Time taken by current to change from 0 to rms value
= Time taken by the voltage to change from 0 to rms value
= Time taken by voltage to change from rms value to 0.
So t = 2 x 10‐3 s as above.
[0.5 mark for correct reasoning & 0.5 mark for correct final result]
Q.112 In an LC circuit, the current is maximum when impedance Z is minimum,
that is, XL= XC
2
[0.5 mark for correct identification of a condition of resonance]
ωL = 1/Cω
L = 1/Cω2
Here C = 100 x 0.01 x 10‐6 F = 10‐6 F
ω = 2π x 8000 rad/s
L = 1/[10‐6 x (2π x 8000)2]
Solving,
L = 125/32π2 mH
[0.5 mark for correct value
substitution]
[1 mark for correct calculation]
Competency Focused Practice Questions | Physics | Grade 12
90
Q.113 When dc is applied:
2
ω = 0, and Z = R
I = V/R
R = V/I = 50 / 0.5 = 100 ohm
[0.5 mark for correct value of R]
When ac is applied:
I = V/Z
Z = V/I = 50/0.2 = 250 ohm
[0.5 mark for correct value of Z].
As Z2 = R2 + ω2L2
ω2L2 = Z2 ‐ R2
= (250)2 – (100)2 = 150 x 350
Solving for L,
L = √21/2π H
[1 mark for correct calculation of L]
Q.114 The time taken for half a cycle of sinusoidal variation of current corresponds to
a pair of successive magnetization and demagnetization of the inductor.
2
[0.5 mark for identification of the relation between the magnetization cycle of
the inductor and the sinusoidal variation of the current]
So,
8 x 10‐3s
= half the time period of the ac cycle
So T/2 = 8 x 10‐3
T = 16 x 10‐3s
[1 mark for correct calculation of time period]
Frequency ν = 1/T = 103/16 = 62.5 Hz
[0.5 mark for correct calculation of frequency]
Competency Focused Practice Questions | Physics | Grade 12
91
Q.115 Resistance R of the bulb, P = V2/R
2
R = V2/P = 100 x 100/50 = 200 ohm
[0.5 mark for correct value of R]
With the choke coil in series to the bulb,
the current in the circuit,
[0.5 mark for correct equation of I]
Voltage drop across the bulb must be 100 V, So 100= IR
Transposing and solving
L2 = 12/π2
L =√12/π H = 1.10 H (approx)
[1 mark for the correct value of L]
Competency Focused Practice Questions | Physics | Grade 12
92
5. ATOMS AND NUCLEI
Q. No
Question
Marks
Multiple Choice Question
Q.116 If an electron in a hydrogen atom undergoes transition from 1st to 3rd energy
level, by what factor does the time period of revolution undergo a change?
A.
remains the same
B.
becomes 3 times the initial value
C.
becomes 9 times the initial value
D.
becomes 27 times the initial value
Q.117 Hydrogen atom at its ground state is excited by incident photons of energy
12.75 eV. What is the expected count of the distinct spectral lines emitted by
this hydrogen atom?
A.
3
B.
5
C.
6
D.
8
Q.118 An electron in the Bohr model atom is in the nth orbit of radius r and has kinetic
energy KE. Which of the following statements does not comply with the
postulate of Bohr’s model of atom?
1
1
1
A. Radius of the orbit in terms of de Broglie wavelength is given as nλ/2π
B. In going from lower to higher orbit, both PE and TE increases but KE
decreases.
C. If KE = 3.4 eV, the corresponding potential energy PE is 6.8 eV and Total
energy TE is 10.2 eV
D. If the ionization potential of the given Bohr model atom is 13.6 V, the
energy required to remove the electron from its second orbit is 3.4 eV.
Competency Focused Practice Questions | Physics | Grade 12
93
Q.119 Choose the only correct statement.
A. In a stable nucleus, the number of neutrons ≥ number of protons.
1
B. Nuclei with atomic number greater than 82 show a tendency to fuse.
C. A stable nucleus in general has even number of protons and odd number
of neutrons.
D. The proton‐proton nuclear force > proton‐neutron nuclear force > neutron‐
neutron nuclear force.
Q.120 Binding energy values of a few nuclei are given in the table below.
Lithium‐7
Iron ‐56
Lead ‐ 206
Deuterium ‐2
40.1 MeV
492.2 MeV
1622 MeV
2.2 MeV
1
The least and the most stable nuclei respectively are:
A.
Lead and Lithium
B.
Lithium and Deuterium
C.
Deuterium and Iron
D.
Iron and Lead
Q.121 Study the following statements carefully and identify the correct statement/s.
1
I. The nuclear pairs, 7N14 and 6C13 & 15P30 and 14Si30, both constitute isobars.
II. 7N14 can absorb a neutron and transform into 3Li7 after emitting 2 alpha
particles and a proton.
III. If 5.6 MeV is required to remove 1 neutron from 3Li7 and transform it to 3Li6,
then it is also the BE per nucleon of 3Li6
IV. If M is the mass of the nucleus 13Al27, and mp & mn are the masses of
protons and neutrons respectively, the binding energy of the nucleus is [|M ‐
13mp ‐ 14mn |c2 ]
A.
Only statement III
B.
Only statements I and III
C.
Only statements II and IV
D.
Only statements III and IV
Competency Focused Practice Questions | Physics | Grade 12
94
Q.122 Which of the following statements are INCORRECT for atomic nuclei?
A. As mass number A increases, the density of the nucleus does not change.
1
B. Heavier nuclei tend to have larger Z/N ratio as Coulomb forces have longer
range compared to nuclear forces.
C. For nuclei with A > 100, the binding energy per nucleon of the nucleus
decreases on an average as A increases.
D. Two Lithium nuclei do not combine to form carbon nuclei at room
temperature because Coulomb repulsions do not allow them to come very
close.
Free Response Question / Subjective Question
Q.123 Particles like electron, protons and alpha particles can be used to ionize
hydrogen atom. If the ionization energy of hydrogen atom is Ei and all these
striking particles carry same kinetic energies, then determine which of these
particles will be most effective in ionizing the hydrogen atom?
3
Assume that the collisions are perfectly inelastic and it’s the excess KE that is
used for ionisation.
Q.124 An ultraviolet radiation falls on a H atom.
2
(a) If the H atom is in the ground state when the incident photon of wavelength
80 nm falls on it, will the H atom get excited to the higher energy levels? OR will
the H atom
get ionized?
Justify your answer.
(b) Post your answer of part (a),
Determine EITHER the quantum number of the energy level to which the H
atom gets excited OR determine the KE of the electron emitted out of the H
atom.
Use value of hc = 1240 eV‐nm
Competency Focused Practice Questions | Physics | Grade 12
95
Q.125 (a) Show that it takes more energy, in general, to remove a neutron from a
nucleus than a proton. Use the example of ZAX and the below data for the
3
purpose‐of‐explanation.
Mass‐of ZAX=63.546u
Massproton =1.0073u
Massneutron =1.0087u
Massof ZA‐1X=62.5493u
Mass of Z‐1A‐1Y = 62.5467 u (b) Give reason: Why does it take more energy to
separate a neutron from the nucleus than a proton?
Competency Focused Practice Questions | Physics | Grade 12
96
Answer Key & Marking Scheme
Q.No
Answers
Marks
Q.116 D. becomes 27 times the initial value
1
Q.117 C. 6
1
Q.118 C. If KE = 3.4 eV, the corresponding potential energy PE is 6.8 eV and Total energy
TE is 10.2 eV
1
Q.119 A. In a stable nucleus, the number of neutrons ≥ number of protons.
1
Q.120 C. Deuterium and Iron
1
Q.121 D. only statements III and IV
1
Q.122 B. Heavier nuclei tend to have larger Z/N ratio as Coulomb forces have longer
range compared to nuclear forces.
1
Q.123 The momentum conservation during the collision:
3
mu = (m + mH)v
where m is the mass of the ionizing particle and mH is mass of hydrogen atom
Initial KE, Ki = (1/2) mu2 Final KE, Kf = (1/2) (m+mH)v2
[0.5 mark for each of initial and final KE]
The decrease in KE is used for ionization of the H atom So (1/2) mu2‐
(1/2)(m+mH)v2 = Ei
Substituting for v = mu/(m+mH ) and transposing,
Ei = [mH/(m+mH )] Ki
[1 mark for correct expression for ionisation energy]
Since Ki is same for all types of striking particles, the particles with minimum
mass will be most effective in ionizing the Hydrogen atom.
Amongst the given list, electrons having minimum mass will be most effective.
[1 mark for the correct conclusion and reason]
Competency Focused Practice Questions | Physics | Grade 12
97
Q.124 (a) Energy of the incident 80 nm photon E = hc/λ = 1240/80 = 15.5 eV As the
ionization energy of the H atom is +13.6eV, it is less than the energy 15.5 eV
absorbed from the incident photon, so the H atom gets ionized.
2
[0.5] mark for the calculation of energy of incident photon]
[0.5 mark for the correct conclusion]
(b) KE of the emitted electron = Energy of the incident 80 nm photon – Ionization
energy = 15.5 – 13.6 = 1.9 eV
[1 mark for the correct calculation of the KE of the emitted electron]
Q.125 (a) Removing a neutron from ZAX :
3
ZAX ‐> ZA‐1X + 01n
The binding energy of the missing neutron = mass defect x 931 MeV
Mass defect = [Mass of ZA‐1X + Massneutron ] ‐ Mass of ZAX
= 62.5493 u + 1.0087 u ‐ 63.546 u= 0.012 u
BE = 0.012 x 931 = 11.172 MeV
[0.5 mark for the correct calculation of mass defect]
[0.5 mark for the correct calculation of the BE]
Removing a proton from ZAX :
ZAX ‐> Z‐1A‐1Y + 11p
The binding energy of the missing proton = mass defect x 931 MeV
Mass defect = [Mass of Z‐1A‐1Y + Massproton ] ‐ Mass of ZAX
= 62.5467 u + 1.0073 u ‐ 63.546 u= 0.008u
BE = 0.008 x 931 = 7.48 MeV
Hence BE to remove a neutron is more than that of the proton.
[0.5 mark for the correct calculation of mass defect]
[0.5 mark for the correct calculation of the BE]
b) Protons are bounded by attractive nuclear forces and also experience
repulsion due to electrostatic forces.
Neutrons are bounded by attractive nuclear forces only. Hence they require
greater energy to get separated from the nucleus.
[0.5 mark for each correct statement of p and n]
Competency Focused Practice Questions | Physics | Grade 12
98
6. DUAL NATURE OF RADIATION AND MATTER
Q.No
Question
Marks
Multiple Choice Question
Q.126
Q.127
Select the correct statement regarding photoelectric emission.
A.
The number of photoelectrons is proportional to the frequency of the
light.
B.
The work function of the metal varies as a function of the depth from the
surface.
C.
Photoelectrons are ejected from the metals only if the incident radiation
has a certain minimum wavelength.
D.
The photoelectrons produced by a monochromatic incident light incident
on the metal surface have a spread of kinetic energies.
ASSERTION: A photon is different from a normal particle.
1
1
REASON 1: Energy of the photon in motion is purely kinetic as it has no rest
energy. REASON 2: A photon travels with the speed of light in vacuum.
REASON3: Energy of the photon at rest becomes purely potential as its rest
mass is zero. Select the correct option.
Q.128
A.
Assertion is correct. Only reason 3 is correct.
B.
Assertion is incorrect. Only Reason 2 is correct.
C.
Assertion is correct. Only reasons 1 and 2 are correct.
D.
Assertion is incorrect. Only Reasons 1 & 3 are correct.
Light from a source of frequency f > fo (threshold frequency) is incident on a 1
metal surface. If the intensity of incident radiations from the light source is
increased keeping the frequency f of light constant, which of the following
quantities associated with emitted photoelectrons DOESNOT remain the same?
A.
only photoelectric current
B.
only de Broglie wavelength
C.
both momentum and maximum KE
D.
both de Broglie wavelength and the photo current
Competency Focused Practice Questions | Physics | Grade 12
99
Q.129
Given are the maximum speeds of the photoelectrons corresponding to a
1
wavelength of incident light on a certain metal surface.
Wave length Maximum speed of theemitted electrons
λ1 v1
λ2 v2
If v2 < v1, then, which of the following is/are CORRECT?
I. Wavelength λ1 > Wavelength λ2
II. Frequency f1 < Frequency f2
III. Maximum KE due to wavelength λ1 > Maximum KE due to wavelength λ2
A.
All are correct
B.
Only III is correct
C.
Only I and II are correct
D.
Only II and III are correct
Free Response Question / Subjective Question
Q.130
Given are the work functions of two materials.
3
Material
Work function (eV)
Sodium
2.4
Zinc
4.3
A radiation of wavelength 400 nm is made to fall on each of the above
materials. In which of the two materials is the photoelectric emission observed?
Take h = 4.1 × 10‐15 eV‐s
Q.131
A light of wavelength 330 nm causes photoelectric emission from a metal of 2
work function 1 eV. What value of collector potential would stop the
photoemission? Take hc = 1240 eV‐ nm
Competency Focused Practice Questions | Physics | Grade 12
100
Q.132
A light of wavelength 3000 Angstrom emits photoelectrons of maximum KE of
0.5 eV.
2
What will be the maximum KE of photoelectrons if the wavelength is changed
to 2000 Angstrom? Take hc = 12400 eV‐Å
Q.133
A powerful UV laser emits radiation of wavelength 300 nm with a power of 5 2
mW. Determine the number of photons of this beam that strikes a metal
surface every second. Take hc ~ 2 x 10‐16 J‐nm
Competency Focused Practice Questions | Physics | Grade 12
101
Answer Key & Marking Scheme
Q. No
Answers
Marks
Q.126 D. The photoelectrons produced by a monochromatic incident light incident on
the metal surface have a spread of kinetic energies.
1
Q.127 C. Assertion is correct. Only reasons 1 and 2 are correct.
1
Q.128 A. only photoelectric current
1
Q.129 B. Only III is correct
1
Q.130 Threshold wavelength,
3
[1 mark for the correct formula for threshold wavelength]
For sodium,
= 5.12 x 10‐7 m = 512 nm
Since the λincident < λthreshold ,
there will be a photoelectric emission observed in sodium.
[1 mark for the correct calculation and final result] For Zinc,
= 286 nm
Since the λincident > λthreshold,
there will be no photoelectric emission observed in zinc. [1 mark for the correct
calculation and final result]
Competency Focused Practice Questions | Physics | Grade 12
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Q.131 Equation:
2
Φo = hν ‐ KEmax = (h c/λ) ‐ KEmax
Here KEmax = eVo where Vo is the stopping potential applied to the collector
plate
Φo = (h c/λ) ‐ eVo
[1 mark for the correct relation between work function and collector stopping
potential]
[1 mark for the correct calculation ofthe collector potential ]
Q.132 Using
2
In the first case,
Φo = (12400/3000) – 0.5
In the second case,
Φo = (12400/2000) – K
[0.5 mark for each equation] Equating and solving for K
We get
K = 2.57 eV for wavelength 2000 Å
[1 mark for correct final result]
Q.133 Energy of each photon = hν Power of the beam = n hν = n hc/λ
[1 mark for the correct formulae of Energy and power in terms of frequency]
2
5 x 103 = n x 2 x 1016
300
Upon calculating,
n = 7.5 x 1015 photons per second [1 mark for the correct final result]
Competency Focused Practice Questions | Physics | Grade 12
103
7. ELECTROMAGNETIC WAVES
Q. No
Question
Marks
Multiple Choice Question
Q.134 Study the following statements carefully.
A.
Electric and magnetic fields have zero average value in a plane em wave.
B.
For an em wave, the ratio k/ω is independent of wavelength.
C.
In an em wave, the E and B fields vary with the same frequency and are in
opposite phase.
D.
Since E = cB, the energy associated with the electric field is much greater
than that associated with the magnetic field.
1
Identify the correct option.
A.
only A and B are correct
B.
only C and D are correct
C.
only A and C are correct
D.
only B and D are correct
Q.135 Which of the following statement/s are incorrect?
A.
The displacement current flows in a dielectric of the capacitor when the
potential difference across its plates is decreasing with time.
B.
The direction of propagation of electromagnetic waves is given by
C.
The dimensions of
D.
Instantaneous energy flow rate is a constant for an electromagnetic wave.
E.
Light of uniform intensity shines perpendicularly on a totally absorbing
surface.
1
are the same as that of electric voltage.
On decreasing the area of the surface, the intensity remains the same.
A.
Only statements A & B
B.
Only statements C & D
C.
Only statements D & E
D.
Only statements A, C & D
Competency Focused Practice Questions | Physics | Grade 12
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Q.136 The diagrams below show the electric and magnetic field components of an
electromagnetic wave at a certain time and location.
1
Which of these electromagnetic waves are travelling towards you?
‘
A.
only the em wave in Fig I
B.
only the em wave in Fig I and II
C.
only the em wave in Fig II and III
D.
only the em wave in Fig II, III and IV
Free Response Question / Subjective Question
Q.137 An electromagnetic wave of frequency 1 GHz travels through an empty space
along the z‐direction. At a specific point in space, the electric field E attains a
maximum value of 50 V/m. If the electric wave is polarized along x‐axis, then,
3
(a) explain and identify the plane in which the magnetic field B will lie. (b)
express the electric and magnetic fields as a function ofz and t.
Q.138 In a spaceship orbiting around the Earth, two astronauts stationed 2 m apart are
speaking to each other. The conversation is transmitted to Earth via
electromagnetic waves.
2
Given that the sound waves travel through the air between the two astronauts
in exactly the same time as the em waves take to reach the Earth ground station.
Calculate the distance of the spaceship from the ground station. Take speed of
sound in the air between the two astronauts as 340 m/s.
Competency Focused Practice Questions | Physics | Grade 12
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Q.139 An unfortunate nuclear explosion leaves behind the residual gamma radiations
in the vicinity of the explosion site with an average energy density of 4 × 10−14
J/m3.
2
(a) What is the rms value of the electric field of the radiation?
(b) Compare the electric field strength with the magnetic field strength in this
residual radiation.
Q.140 A dish antenna with a circular aperture of a radius 20 cm, receives digital TV
signals from a satellite. The average intensity of the em waves that carry a
particular TV program is 5 x 10‐14 W/m2.
2
Determine the following:
(a) electromagnetic energy delivered to the dish during the telecast of 30
minutes of a programme.
(b) average energy density of the em wave.
Q.141 A laser emits a sinusoidal em wave of wavelength 10 μm along the x‐axis. The E
field of the wave is parallel to the –ve z‐axis with a maximum value of 1 MV/m.
3
Express the wave equation for E and B for this wave with all appropriate values
and directions.
Q.142 A satellite at a height of 100 km from the Earth’s surface detects a radio signal
emitted by a radio station on the ground. If the average power of the signal
received is 100 kW, find the amplitudes E0 and B0 of the incoming signal.
Competency Focused Practice Questions | Physics | Grade 12
3
106
Answer Key & Marking Scheme
Q. No
Answers
Marks
Q.134 A. only A and B are correct
1
Q.135 B. Only statements C & D
1
Q.136 D. only the em wave in Fig II, III and IV
1
Q.137 (a) Since E wave is polarized along x‐direction and the em wave propagates along
z‐direction, the magnetic field vector has to be perpendicular to both E wave
and the direction of propagation of the wave. So, B vector is aligned along y axis
and lies in y‐z plane.
3
[1 mark for correct explanation and the direction]
(b) The standard waveforms of E and B vector in an em wave are: E = Eo sin (kz
‐ ωt)
B = Bo sin (kz ‐ ωt)
Bo = Eo/c = 50/c T
k = 2π/λ = 2π ν/c = 2π x 109/c
E = 50 sin (2π x 109 z ‐ 2π x 109.t)
C
E = 50 sin [2π x 109. (z/c ‐ t)] B = (50/c).sin [2π x 109. (z/c ‐ t)]
[1 mark for each correct final equation of E and B]
Q.138 For the travel of sound waves between the two astronauts: 2 / 340 = t ….(1)
2
For the travel of em waves from the spaceship to the Earth station: d/(3 x 108)
= t ….(2)
[0.5 mark for each of the equations for sound and em waves] Equating (1) and
(2) and solving for d,
d ≈ 1765 km
[1 mark for correct final result]
Competency Focused Practice Questions | Physics | Grade 12
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Q.139 a. As energy density u = ε0E2
2
= 0.067 N/C
[1 mark for the correct finalresult]
b. In any em radiation, the ratio E/B = c, is always constant [1 mark for the
correct application of the ratio between E to B in any em wave]
2
Q.140 (a) Average intensity I = average power P /area A Average power P = I. A = I. πr
Average energy delivered during the telecast = I. πr2. t
2
[0.5 mark for correct formula of energy in terms of intensity, area and time]
E = 5 x 10‐14 x π x (0.2)2x 1800
= 11.3 x 10‐12 J
[0.5 mark for correct value of energy]
(b) Energy density u = I/c = 5 x 10‐14 / 3 x 108 = 1.66 x 10‐22 J/m3[1 mark for
correct result of energy density]
Q.141 The standard wave equations:
3
E‐z = Eo sin(kx ‐ ωt) … direction of E field will be along –z‐axis By = Bo sin(kx ‐
ωt) .. direction of B field will be along y‐axis
Where Bo = Eo/c = 106/3 x 108 = 3.3 x 10‐3 T k = 2π/λ = 2 π / 10 x 10‐6 = 2π x 105
/m
ω = ck = 3 x 108x 2π x 105 = 6π x 1013 rad/s
[0.5 mark for correct representation of E and correct values of Bo, k and ω]
Equations:
E‐z = ‐ k (106 V/m) sin(2π x 105. x ‐ 6π x 1013 .t)
By = j (3.3 x 10‐3 T) sin(2π x 105. x ‐ 6π x 1013 .t) [0.5 mark for each of E and B
equations]
Q.142 Surface area of the hemisphere on the ground through which the signal is
emitted by the radio station
A = 2πR2 = 2π (100 x 1000)2 = 2π x 1010 m
[0.5 mark for correct calculation of surface area]
Competency Focused Practice Questions | Physics | Grade 12
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Intensity of the signal received by the satellite = I = Average power/ Area = 100
x 1000 / 2π x 1010 = 10‐5/2π W/m2
[1 mark for correct calculation of Intensity of signal]
As I = ∈o E2rms c
Eo = 0.034 V/m
[1 mark for correct calculation of Eo]
Bo = Eo/c = 0.0115 x 10‐8 T
[0.5 mark for correct calculation of Bo]
Competency Focused Practice Questions | Physics | Grade 12
109
8. OPTICS
Q.No
Question
Marks
Multiple Choice Question
Q.143 An object is moved towards a concave mirror at a constant speed, from infinity 1
to its focus. Which of the following statements correctly describe the
corresponding motion in the image formed by the concave mirror?
A. The image moves slower initially and faster later on, away from the mirror.
B.
The image moves faster initially and slower later on, towards the mirror.
C.
The image moves at a constant speed, faster than the object, away from the
mirror.
D. The image moves at a constant speed, slower than the object, towards the
mirror.
Q.144 Read the statements given here carefully. Each statement is in reference to 1
spherical mirrors. Here u, v and f denote the object distance, the image distance
and the focal length of the spherical mirror respectively.
I.
A graph plotted between u and v is a hyperbola and when u = f, v = Infinity.
II.
A graph plotted between u and v is a straight line and when u = Infinity, v =
f.
III. A graph plotted between (1/u) and (1/v) is a straight line with an intercept
of each axis as (1/f).
IV. A graph plotted between (1/u) and (1/v) is a hyperbola with an intercept of
each axis as f.
Select the correct option.
A. Statements I and III are correct
B. Statements II and IV are correct
C.
Statements I and II are correct
D. Statements III and VI are correct
Q.145 Read the following statements carefully:
I.
A drop of an oil in water or in a glass, both behave as convergent lens.
II.
A water drop in air and a glass sphere in water, both behave as convergent
Competency Focused Practice Questions | Physics | Grade 12
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110
lens.
III. An air bubble in water and a water bubble in glass, both behave as a
divergent lens.
IV. A frozen ice crystal inside a glass sphere and a bromine liquid drop inside a
glass sphere, behave as divergent lens.
[Reference values of refractive indices of common substances: Air = 1.001;
Water = 1.33; ice = 1.31; Glass = 1.51; Oil = 1.4; Bromine = 1.66]
Select the correct option.
A.
Statements I and II are correct.
B.
Statements I and III are correct.
C.
Statements II and III are correct.
D.
Statements II and IV are correct.
Q.146 An incident light ray falls on a glass prism at an angle of 60o and emerges with
an angle of 30o with its initial incident direction.
1
If the angle of the prism is 30o, then which of the following is an INCORRECT
statement?
A. Refractive index of the prism is √3.
B. The light undergoes minimum deviation through the prism (i.e., r1 = r2).
C.
The emergent ray is perpendicular to the face from which it emerges.
D. Angle of refraction r1 at the incident face is same as angle of the prism.
Q.147 Which of the following actions will lead to an increase in the magnifying power
of an astronomical telescope?
1
A. Increase in the length of the telescope tube.
B. Interchange the objective and the eyepiece of the telescope.
C.
A small piece of paper on the objective of the telescope pointed towards
the moon.
D. Increase in the focal length of the objective and decrease in the focal length
of the eye piece.
Q.148 In a glass prism of refractive index n, identify the condition for which an incident
light emerges out of the prism with a non‐zero deviation.
1
A. Angle of prism is zero
Competency Focused Practice Questions | Physics | Grade 12
111
B
Angle of incidence is equal to angle of emergence
C.
The n of the material of the prism is same as its surroundings
D. Sum of angle of incidence and angle of emergence is equal to angle of prism
Q.149 Study the data related to refractive index and critical angle as given here.
Glass‐air
Glass‐water
Water‐
air
Diamond‐air
Refractive
index, nGA
= 1.5
Refractive index,
nGW = 1.125
Refractiveindex,
Refractiveindex,
Critical
angle, θc =
sin‐
1[1/1.5]=
42o
Critical angle, θc =
sin‐1[1/1.125]=
63o
Critical angle, θc = sin
nWA = 1.33
1[1/1.33]= 49o
1
nDA = 2.5
Critical angle, θc
=sin‐ 1[1/2.5]=
24o
Identify an INCORRECT conclusion.
A.
Lesser is the value of n, greater is the critical angle.
B.
Greater is the wavelength of the incident light, greater is the corresponding
critical angle in a given pair of media.
C.
For given beam of incident light containing all angles of incidences,
maximum total internal reflection is observed at glass‐water interface.
D.
Incident light with angle of incidence equal to 50oin the denser medium
will be total internally reflected in all the above media except on glass‐
water interface.
Q.150 A real image of size p times the size of an object is formed by a concave mirror
of focal length f.
1
What is the object distance from the mirror?
A. (p+1)*f/p
B. (p‐1)*f/p
C. p*f/(p+1)
D. p*f/(p‐1)
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112
Q.151 Two superimposing waves are of amplitudes a1 and a2 and intensities are I1 andI2 1
respectively. If the ratio I1:I2 is 1:16, what will be the ratio of their maximum to
minimum intensities upon super‐imposition?
A. 1/4
B. 4/1
C. 25/9
D. 9/25
Q.152 Two waves from a pair of coherent sources strike the screen at a point. For which 1
of the following cases will the intensity of superimposed waves be maximum
and hence resulting in constructive interference at the given point of
observation?
Path
difference
3λ
Path
Path
Phase
Phase
Phase
difference
difference
difference
difference
difference
3λ/2
4λ/2
‐2π
+5π
0
A. Path difference = 3λ/2; 4λ/2 & Phase difference = 5π
B. Path difference = 4λ/2 & Phase difference = 0,+5π
C.
Path difference = 3λ; 3λ/2 & Phase difference = ‐2π, 5π
D. Path difference = 3λ; 4λ/2 & Phase difference = 0, ‐2π
Q.153 A 5% change in wavelength is observed in the light received from a distant star. 1
What is the speed of the moving star?
A. c
B. 0.05 c
C.
20.0 c
D. 0.02 c
Q.154 Which of the following statements DOES NOT correctly comply with Huygens ’
Principle of constructing a secondary wavefront from a primary wavefront?
1
A. After some time interval, the new position of the wave front is the surface
Competency Focused Practice Questions | Physics | Grade 12
113
tangent to the secondary wavelets.
B.
Secondary wavelets propagate outward through a medium with speeds
characteristic of waves in that medium.
C.
A secondary wavefront is always a plane wavefront irrespective of whether
the primary wavefront is planar or spherical.
D. All points on a given primary wave front are taken as point sources for the
production of spherical secondary waves, called wavelets.
Q.155 Which of the following statements related to forming of interference patterns
are correct?
I.
The light‐wave interference pattern exists in three dimensions.
II.
There is a net energy flow from the slits to the screen in an interference
pattern.
1
III. The light waves from the two slits overlap as they spread out and produce
interference fringes on a screen placed opposite to the slits.
IV. Two different but identical powered white light bulbs placed side by side
would fail to produce an interference pattern on the screen.
A. All are correct
B. Only I and II are correct
C.
Only I, II and III are correct
D. Only II, III and IV are correct
Q.156 When coherent light waves interfere to produce alternate bands of dark and 1
bright interference bands, which of the following statement/s correctly identify
the energy and intensity distribution across the interference bands?
(I) Energy conservation is violated because energy disappears in the dark
bands.
(II) Intensity at the bright bands is four times the square of the amplitude of the
individual waves.
(III) The total energy leaving the slits is distributed among bright and dark bands
and energy is conserved.
(IV) Energy transferred by the light sources at the bright bands is same as carried
by each of the individual waves.
A. Statement I and II only
Competency Focused Practice Questions | Physics | Grade 12
114
B. Statement I and IV only
C. Statement II and III only
D. Statement II and IV only
Q.157 Loud music being played on the TV in the adjacent room can be heard through 1
the slightly open door of your study room. But you cannot see what is happening
outside your room. Why is it that you can hear the sounds but unable to see
through the narrow opening of the door?
A. Sound waves can pass right through the walls, but light waves cannot.
B.
The open door is a small slit for sound waves, but a large slit for light waves.
C.
Light waves diffract efficiently through the single slit of the open doorway,
but sound waves cannot.
D. Light waves can travel only along straight lines whereas sound waves can
travel along curved paths.
Free Response Question / Subjective Question
Q.158 A yellow light of wavelength 6000 Å enters from air into water. Refractive index 5
for water is n = 4/3.
(a) Which of the following quantities for the yellow light change/s? velocity,
wavelength & frequency.
(b) What colour does the yellow light change into upon entering water? Explain
your choice of the colour.
(c) Study each of the characteristics of incident and refracted yellow ray
carefully. Identify if the individual combination of values (v, λ, ν and the colour)
are possible for the yellow light in air undergoing refraction as it enters optically
denser media.
(i) Incident Light in
air
(ii) Refracted
light in media 1
(iii) Refracted light in
media 2
Speed, v
m/s
3 x 108
Wavelength,
λÅ
6000
Frequency,
ν Hz
~5 x 1014
Colour
2.25 x
108
2 x 108
4500
~5 x 1014
Yellow
7000
~5 x 1014
Red
Yellow
Explain your answer in each of the cases.
Competency Focused Practice Questions | Physics | Grade 12
115
Q.159 An object is placed on the optical bench at a distance of 40 cm from the convex
mirror of focal length 10 cm. A plane mirror is placed in between the object and
the convex mirror such that it covers the lower half of the convex mirror.
2
Find the distance from the object where the plane mirror be placed, so that no
parallax is observed between the two images formed by the two mirrors.
Q.160 A beam of yellow light falls with a certain angle of incidence at the interface of
the two media and experiences an angle of refraction of 90o. Red and blue light
beams then replace the yellow light successively, keeping all other conditions
same.
3
One of the two colours, red or blue, undergoes total internal reflection. (a)
Identify the colour of light that undergoes TIR. Give reason for your choice. (b)
Also give reason why the other colour doesn’t undergo TIR.
Q.161 A thin convex and a thin concave lens are placed along the same axis on an
optical bench. A parallel beam of light falls on the convex lens. The distance d
between the two lens is adjusted by moving the concave lens so that the light
emerging out of the concave lens is parallel.
2
(a) Draw a suitable ray diagram to represent the flow of light through the lens
system.
(b) State the condition that ensures that the emergent beam is parallel. If the
focal length (convex) = 20 cm and focal length (concave) = 8 cm, find d.
Q.162 Lens Q when placed in contact with a converging lens P of focal length = 20 cm
makes a combination that behaves as a converging lens system of focal length
30 cm.
3
Lens Q when placed in contact with another lens R makes a combination that
behaves as a diverging lens system of focal length 10 cm.
Identify the nature of lens Q and R and determine their focal lengths.
Q.163 An object placed at a distance of x from the first focus of a convex lens forms a
real image at a distance of y from its second focus.
2
Show that the product xy is equal to the square of the focal length f of the lens.
Q.164 A beam of light consisting of violet (V), green (G) and orange (O) colours is
incident on a right‐angled prism as shown.
Competency Focused Practice Questions | Physics | Grade 12
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116
If the refractive indices of the material of the prism with respect to these three
colours are 1.55, 1.45 and 1.35 respectively, which of these colours are able pass
through the face AC?
Q.165 A light ray entering a right‐angled prism undergoes refraction at the face AC as
shown. (Fig I)
3
(a) What is the refractive index of the prism in Fig I?
Fig I
(b) i. If the side AC of the above prism is now bounded by a liquid of refractive
index 2/√3, (Fig II), determine if the light ray continues to graze along the
interface AC OR undergoes total internal reflection OR undergoes refraction into
the liquid.
Competency Focused Practice Questions | Physics | Grade 12
117
Fig II
ii. Draw the ray diagram to represent the path followed by the incident ray with
the corresponding angle values.
Given sin‐1 √2/√3 = 54.6
Q.166 A lens made of glass (nG) is placed in a medium of refractive index (nM).
2
(a) Show that for nM < nG , the nature of the lens remains unchanged but its
focal length increases.
(b) Show that for nM > nG , the nature of the lens changes (i.e., convergent lens
behaves as divergent and vice versa).
Q.167 An object is placed at a distance of 10 cm from first focus of a biconvex lens of 2
focal length f. It forms an image at a distance of 10 cm from its second focus on
the other side of the lens.
Determine the focal length f of the given biconvex lens.
Q.168 In a Young’s double slit experiment arrangement, each slit is illuminated with a 2
light containing two wavelengths, 5890 Å and 5895 Å. Due to the dual
wavelengths, the fringe patterns due to individual wavelengths overlap on the
screen.
Determine up to what order the fringes can be seen on the screen.
(Note: Here the order of the fringes implies the nth fringe of each wavelength
beyond which the interference patterns appear as continuous).
Q.169 A monochromatic light produces a fringe pattern on the screen in a double slit 2
experimental set up. As the screen is moved towards the slits through a distance
of 0.05 m, the fringe width on the screen changes by 3 x 10‐ 5 m.
Competency Focused Practice Questions | Physics | Grade 12
118
a. Does this change represent an increase or decrease in the fringe width of the
pattern?
b. Keeping the wavelength and separation between the slit’s constant,
determine the change in fringe width produced in case the screen is moved
through a distance of 0.02 m from its initial position.
Q.170 Fringe width of the interference pattern on the screen in a double slit
experiment is β.
2
Determine the fringe width if the whole apparatus is immersed in a liquid of
refractive index n = 6/5.
Q.171 A visible light source constituted of two wavelengths: λ = 520 nm and λ' = 420 2
nm is used in a double‐slit interference experiment. Also given are distance
between slits and the screen, D = 1.50 m and the slit width d = 0.025 mm.
Find the order n of maxima of light of λ that will overlap with n' of light of λ' on
the screen.
Q.172 A light of wavelength 500 nm is used to illuminate a double slit experiment. For 2
a double slit with a separation of 0.4 mm, determine how far the screen should
be placed from the slits in order that a dark fringe appears directly opposite to
both the slits and only one bright fringe in between them.
Q.173 A coherent light beam of 500 nm wavelength strikes two slits, producing an 2
interference pattern. The two slits are separated by 0.3 mm and the distance
between the slits and the screen is 0.3 m.
Determine the number of maxima observed across the perpendicular distance
of y = ‐1 mm to y = +1 mm on the screen.
Q.174 In the double slit experiment, given that, d = 0.140 mm, D = 140 cm, λ = 600 nm,
and y = 1.80 cm.
3
(a) Determine the path difference δ for the rays from the two slits arriving at P.
(b) Express this path difference in terms of wavelength λ.
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(c) Does the position P correspond to maxima, minima, or an intermediate
condition?
Q.175 Two cars following each other enter a short tunnel. Car 1 plays music on FM 2
station at frequency 100 MHz whereas Car 2 plays music on AM station at
frequency 1000 kHz. Which radio station will continue to play the music while
the car passes through the short tunnel? Explain your answer.
Q.176 A blue coloured alternate bands of diffraction pattern appears on a screen due 1
to the blue light of wavelength λ passing through a single slit of width w. If the
blue light is replaced with a green light of wavelength 1.5 λ, to what width should
you change the slit in order to get the original pattern back?
Q.177 Sea waves of wavelengths 25 m keep splashing at a 60 m wide opening in the 2
wall barrier near the shore. The racing boats are parked on the other side of the
wall barrier.
a. Determine two consecutive angles to the incident direction at which the boats
can be parked, so that they experience minimum rocking motion.
b. Which of these two angles will receive least disturbance?
Q.178 Monochromatic light produces a diffraction pattern through slit of width 1 mm 2
on a screen that is 2 m away from the slits. If the width of central maxima in the
diffraction pattern is 2 cm, determine the wavelength of the incident
monochromatic light.
Q.179 A combination of multiple convex lens kept in contact with each other has an 3
equivalent focal length of 0.02 m. An object is placed at a distance of 0.03 m
from the combination lens system. If one of the component lens of focal length
0.1 m is removed from the combination, by what distance is the image of the
object shifted from its initial position?
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120
Answer Key & Marking Scheme
Q.No
Answers
Marks
Q.143 A. The image moves slower initially and faster later on, away from the mirror.
1
Q.144 A. Statements I and III are correct
1
Q.145 C. Statements II and III are correct.
1
Q.146 B. The light undergoes minimum deviation through the prism (i.e., r1 = r2).
1
Q.147 D. Increase in the focal length of the objective and decrease in the focal length
of the eye piece.
1
Q.148 B. Angle of incidence is equal to angle of emergence
1
Q.149 C. For given beam of incident light containing all angles of incidences, maximum
total internal reflection is observed at glass‐water interface.
1
Q.150 A. (p+1)*f/p
1
Q.151 C. 25/9
1
Q.152 D. Path difference = 3λ; 4λ/2 & Phase difference = 0, ‐2π
1
Q.153 B. 0.05 c
1
Q.154 C. A secondary wavefront is always a plane wavefront irrespective of whether
the primary wavefront is planar or spherical.
1
Q.155 A. All are correct
1
Q.156 C. Statement II and III only
1
Q.157
1
B. The open door is a small slit for sound waves, but a large slit for light waves.
Q.158 (a) velocity& wavelength
5
[1 mark for correct choice of characteristics]
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(b) The colour of the light is function of its frequency and not its wavelength.
Frequency of light does not change with the change in medium. Hence the
yellow light remains yellow in water as well.
[0.5 mark for correct answer and 0.5 mark for the correct explanation]
(c)
Speed
i
Frequency
Hz
Colour
m/s
Wavelength
Å
3 x 108
6000
~5 x 1014
Yellow
Possible
Refracted
light
2.25 x
4500
~5 x 1014
Yellow
Possible
Refracted
light
2 x 108
7000
~5 x 1014
Red
Not
Incident
Light
ii
iii
108
possible
i. Possible
c = νλ = 6000 x 10‐10 x 5 x 1014 = 3 x 108 m/s .. which is correct.
ii. The speed decreases to 2.25 x 108, implies that the light has entered another
medium.
λ / λo = v/c
λo = 6000 x 10‐10 x 2.25 x 108/3 x 108 = 4500 Å
Since the frequency doesn’t change, the colour of the light remains as yellow in
the medium.
So (ii) is possible.
iii. The speed decreases to 2 x 108, implies that the light has entered another
medium.
As λ / λo = v/c, the wavelength cannot increase.
The colour of light does not change with the change in medium. So (iii) is not
possible.
[0.5 mark for each correct identification]
[0.5 mark for each correct corresponding explanation]
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Q.159 Using mirror formula for the convex mirror, 1/f = 1/v + 1/u 1/10 = 1/v + 1/(‐40)
2
Solving for v, v = 8 cm
[0.5 mark for the correct value of image distance]
For no parallax, the position of the image formed by plane mirror should
coincide with position of the image formed by the convex mirror.
[0.5 mark for identifying the condition of parallax] Hence,
Total distance between object and image formed by convex mirror
= 40 + 8 = 48 cm
Plane mirror has to be placed at 48/2 = 24 cm from the object. [1 mark for the
correct final result]
Q.160 (a) Blue colour undergoes TIR.
3
As n is proportional to 1/λ and sin c = 1/n
Since λb < λy, nb > ny
So Critical angle Cb < Cy
In the present situation, since at the given angle of incidence i, the yellow light
has angle of refraction = 90, at the same angle of incidence i, the blue light will
undergo total internal reflection since Cb < Cy .
[0.5 mark for the correct colour identification] [1 mark for the correct reasoning]
(b) Red colour does not undergo TIR. It passes through the interface with angle
of refraction < 90.
As n is proportional to 1/λ and sin c = 1/n Since λr > λy, nr < ny ,
So critical angle Cr > Cy
In the present situation, since at the given angle i, the yellow light has angle of
refraction = 90, at the same angle of incidence i, the red light simply passes
through into the rarer medium at an angle of refraction < 90.
[0.5 mark for the correct colour identification] [1 mark for the correct reasoning]
Competency Focused Practice Questions | Physics | Grade 12
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Q.161 (a)
2
(b) Parallel beam that enters the convex lens converges at its focus after
refraction. If the emergent beam from the concave lens is to be parallel, then
the rays coming from convex lens must virtually meet at the focus of concave
lens. [1 mark for the correct reasoning]
Distance d = fcx ‐ fcv = 20 – 8 = 12 cm [1 mark for the correct result]
Q.162 P & Q combination:
3
Calculating f2= ‐60 cm
[1 mark for the correct result of focal length of lens Q]
That is, Lens Q is a diverging lens.
[0.5 mark for correct identification of lens Q]
Q & R combination:
Calculating f2 = ‐12 cm
[1 mark for the correct result of focal length of lens R] That is, Lens R is also a
diverging lens.
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[0.5 mark for correct identification of lens R]
Q.163 Using lens formula,
2
[1 mark for correct formula and substitution] f(x + 2f + y) = xy + fy + fx + f2
Transposing and solving xy = f2
[1 mark for correct proof]
Q.164 Angle i at face AB is 0. So each colour passes through AB and strikes the face AC
at an angle of 45.
2
The light ray will pass through AC if, Angle, i < θc
Or get total internally reflected from AC if i ≥ θc
[0.5 mark for correct identification of condition for TIR] sin i ≥ sin θc
sin 45 ≥ sin θc
1/√2 ≥ sin θc 1/√2 ≥ 1/n
n ≥ √2 = 1.41
[0.5 mark for correct value of n]
So, the light colour with n > 1.41, that is, violet and green are total internally
reflected by the face AC.
Red colour however passes through the face AC as it n < 1.41.
[0.5 mark for correct identification of colours that pass through and the colour
that undergo TIR]
Q.165 (a) Since the light ray enters perpendicular to the face AB, the angle of incidence
on face AC will be 45.
3
So,
n = 1/sin θc sin θc = 1/n
sin 45 = 1/n = 1/√2 So n = √2
Competency Focused Practice Questions | Physics | Grade 12
125
[0.5 mark for the correct calculation of value of n]
(b) In Fig II, the face AC of the prism surrounded by a liquid. n = ng /nl = √2 /
(2/√3) = √3/√2
sinθc = 1/n = √2/√3
θc = sin‐1 √2/√3 = 54.6.
Since the angle of incidence on the surface AC is 45, which is less than the critical
angle for the pair of media (glass and the liquid), the ray NEITHER undergoes
grazing along surface AC, NOR does it undergo TIR.
[1 mark for concluding that the ray does not undergo TIR at AC] Instead it
passes through the surface AC and undergoes refraction in the liquid. For
refracting interface AC, n1sini = n2sinr
n2. sin45 = (2/√3) sinr
sinr = √3/2 r = 60.
[1 mark for concluding that the ray undergoes refraction and calculation of
angle of refraction at AC]
[0.5 mark for the correct diagram of the path of the refracted ray]
Q.166 (a) As 1/f = (n‐1) [1/R1 +1/R2]
2
= (n‐1)K
Here n = nG/nM >1
And since 1/f is proportional to (n‐1),
1/f decreases and f(focal length) increases.
And there is no change in the nature of the lens. [1 mark for the correct
Competency Focused Practice Questions | Physics | Grade 12
126
explanation]
(b) Here n = nG/nM < 1 And since
1/f is proportional to (n‐1) ‐‐> f becomes negative
The nature of the lens changes.
A convergent lens behaves as divergent. A divergent lens behaves as
convergent.
[1 mark for the correct explanation]
Q.167 Given,
2
u = (10+f) v = ‐(10+f)
Using lens formula,
1/f = 1/(10+f) ‐ 1/(‐(10+f) )
1/f =2/(10+f)
10 + f = 2f
f = 10 cm
[1 mark for correct formula and use of correct sign conventions] [1 mark for
correct calculation and result]
Q.168 The fringes appear till the point where a maxima of one wavelength coincides
with minima of the other.
2
[0.5 mark for the statement of the correct condition] That is,
5895 x n = 5890 (n + ½ )
[0.5 mark for the statement of the correct equation] Solving, we get, n = 589
So fringes will be visible up to order of 589 of 5895 Å. [1 mark for the correct
final result]
OR
The fringes appear till the point where a maxima of one wavelength coincides
with minima of the other.
[0.5 mark for the statement of the correct condition] 5890 x n = 5895 (n + ½ )
[0.5 mark for the statement of the correct equation]
Solving n = 589.5
Competency Focused Practice Questions | Physics | Grade 12
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So fringes will be visible upto order of 589 of 5890 Å. [1 mark for the correct final
result]
Q.169 a. Decrease in fringe width.
2
[0.5 mark for the correct conclusion]
b. As β = (D/d) λ
Change in fringe width as a function of D,
Δβ = ΔD/d . λ
[0.5 mark for the correct relation between the changes in β and D] In the first
case,
Δβ1 proportional to ΔD1
Δβ2 proportional to ΔD2
Finding the ratio and calculating, Δβ2= (ΔD2)/(ΔD1 ) . Δβ1
= (0.02/0.05) x 3 x 10‐5
= 1.2 x 10‐5 m
[1 mark for the correct final result]
Q.170 As β = (D/d) λ
2
When immersed in a liquid, D and d remain constant.
The wavelength of the light from the source changes to λ’ as the frequency f of
the light remains constant, speed of light and its wavelength change in the
liquid.
[1 mark for the correct statement on changes in λ, v and f due to change in
medium]
λ’ = v/f λ = c/f
So,
β'/β = λ'/λ = v/c = 1/n Hence, β’ = β/n = 5β/6
[1 mark for the correct finalresult]
Q.171 Condition for nth fringe of maxima of light of λ to overlap with n’ th fringe of
light of λ' on the screen:
Competency Focused Practice Questions | Physics | Grade 12
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128
nλD/d =n'λ'D/d
[1 mark for the statemnet of correct condition and its equation] λ/λ'= n'/n=
520/420=26/21
So 26th bright fringe of light of 420 nm overlaps with the 21st bright fringe of
light of 520 nm.
[1 mark for the correct finalresult]
Q.172 For the first dark fringe to appear right in front of the slits,
2
here n = 0, y= 0.2 x 10‐3 m, λ = 500 x 10‐9 m [1 mark for the correct formula and
values]
= 0.32 m = 32 cm
[1 mark for the correct final result]
Q.173 Given D = 0.3 m, d = 0.3 x 10‐3 m; λ = 500 x 10‐9 m
2
To determine the no. of maxima between of y = ‐1 mm to y = +1 mm. As y = n
λD/d
At y = 1 mm = 1 x 10‐3 m,
[1 mark for the correct calculation of n value for the given y]
So, it implies that 2 maxima are formed on the either side of the central maxima.
So the total count of number of maxima across y = ‐1 mm to y = +1 mm on the
screen is 2 + 1 +2 = 5.
[1 mark for the correct final count of the maxima fringes]
Q.174 a. Path difference,
Competency Focused Practice Questions | Physics | Grade 12
3
129
[1 mark for the correct calculation of the path difference] b.
[1 mark for the correct calculation of n as the coefficient of wavelength]
c. Since n is an integer, the position P would correspond to a maxima.
[1 mark for the correct conclusion on the nature of the fringe]
Q.175 FM station radiates waves at wavelength:
2
AM station radiates waves at wavelength:
[0.5 mark for the calculation of each wavelength]
The width of the tunnel is of the order of wavelength of FM waves. So FM waves
can diffract and enter the tunnel.
Thereby the Car 1 continues to play the music.
On the other hand the AM waves are way more than average width of the
tunnel and fail to diffract at the opening of the tunnel.
So, the AM radio signal fades away as soon as the car enters the tunnel and the
music in the Car 2 dies away inside the tunnel.
[0.5 mark for each of the correct explanation]
Q.176 Answer:
1
The angle of diffraction θ depends on the ratio of wavelength to slit width.
sinθ = +/‐ n λ/a
In order to keep the angle same, if wavelength rises by a factor of 1.5, the slit
width also to increase by the same factor of 1.5.
Then the fringe widths of diffraction pattern due to green light will stay same as
that with blue light.
[0.5 mark for the correct formula] [0.5 mark for the correct final result]
Competency Focused Practice Questions | Physics | Grade 12
130
Q.177 a. The sea waves undergo diffraction at the 60 m wide opening.
2
For minimum rocking of the racing boats, they need to be parked along the angle
where the first or second minima of the diffraction pattern are produced.
[0.5 mark for the correct identification of the points where the boats should be
parked for minimum disturbance]
Using the equation sinθ = nλ/a , where n = 1, we can determine the angle for
first minimum:
sin θ = 25/60 = 0.416 θ = sin‐1(0.416)
[0.5 mark for the correct result of the first minimum angle] The angle of the
second minimum,
sinθ = nλ/a , where n = 2,
sinθ= 2 x 25/60 = 0.83 θ = sin‐1(0.83)
[0.5 mark for the correct result of the second minimum angle]
b. At the second minimum, the disturbance on the boats will be negligible.
[0.5mark for the correct conclusion]
Q.178 For small diffraction angles,
2
[1 mark for the correct formula and angle value]
Also using
λ = d x θ = 1 x 10‐3x 0.5 x 10‐2 = 0.5 x 10‐5 m
So λ = 5000 nm
[1 mark for the correct final result]
Competency Focused Practice Questions | Physics | Grade 12
131
Q.179 Initially,
3
1/v1 ‐1/u = 1/F
1/v1 ‐1/(‐3) = 1/2
So initially the image is formed at v1 = +6 cm
[1 mark for finding the initial position of the image]
For the combination lens system,
1/F = 1/f1 + 1/f2 + 1/f3 +⋯. = 1/f1 + 1/F’, here f1 is the focal length of the lens that
is removed from the combination,
1/F' =1/F ‐ 1/f1 = 1/2 ‐ 1/10
F’ = 2.5 cm.
This is focal length of the remaining combination after the removal of one lens.
[1 mark for finding the focal length of the combination after the removal of the
lens]
1/v2 ‐ 1/u = 1/F'
1/v2 ‐ 1/(‐3) = 1/2.5
v2 = + 15 cm
So the final image of the object is shifted away from the lens system by a
distance of 9 cm.
[0.5 mark for finding the image position in the new combination]
[0.5 mark for the correct shift in the position of the image]
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